MC

87d6_161c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

47e4_ae4b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect MC

99b6_a4ea

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

3dda_4e40

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect MC

2ade_53c7

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect MC

7c5b_889d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect MC

f896_359b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect MC

3dda_7087

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect MC

082d_18fd

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly three (3) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
3
(¼)3⋅(¾)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¼)3⋅(¾)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 35 
 4345 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
3
(½)3⋅(½)5 = 
 8! 
 (8–3)! ⋅ 3! 
(½)8 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
3
(¾)3⋅(¼)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¾)3⋅(¼)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4345 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

0019_b787

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

df77_c986

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct MC

f896_16ee

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect MC

5051_3229

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly four (4) boys ♂ and five (5) girls ♀?

⟮  9  ⟯
4
(½)4⋅(½)5 = 
 9! 
 (9–4)! ⋅ 4! 
(½)9 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 29 
 = 
 126 
 512 
 = 0.2461 = 24.6%
Correct
⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
4
(¾)4⋅(¼)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¾)4⋅(¼)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4445 
 = 
 126×81 
 262144 
 = 0.0389 = 3.9%
Incorrect
⟮  9  ⟯
4
(¼)4⋅(¾)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¼)4⋅(¾)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 35 
 4445 
 = 
 126×243 
 262144 
 = 0.1168 = 11.7%
Incorrect MC

2054_8226

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect MC

cdd7_62d1

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect MC

b96d_5d62

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct MC

8802_ce7c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly six (6) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
6
(¼)6⋅(¾)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¼)6⋅(¾)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 32 
 4642 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
6
(¾)6⋅(¼)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¾)6⋅(¼)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4642 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
6
(½)6⋅(½)2 = 
 8! 
 (8–6)! ⋅ 6! 
(½)8 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

0019_a7a7

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect MC

5051_a7a2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly four (4) boys ♂ and five (5) girls ♀?

⟮  9  ⟯
4
(½)4⋅(½)5 = 
 9! 
 (9–4)! ⋅ 4! 
(½)9 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 29 
 = 
 126 
 512 
 = 0.2461 = 24.6%
Correct
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect
⟮  9  ⟯
4
(¾)4⋅(¼)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¾)4⋅(¼)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4445 
 = 
 126×81 
 262144 
 = 0.0389 = 3.9%
Incorrect
⟮  9  ⟯
4
(¼)4⋅(¾)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¼)4⋅(¾)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 35 
 4445 
 = 
 126×243 
 262144 
 = 0.1168 = 11.7%
Incorrect MC

6037_e55f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct MC

2ade_de7c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect MC

2ade_92cc

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

d2c1_5e22

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect MC

99b6_d068

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect MC

d2c1_7f7b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

6037_2fdb

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect MC

47e4_ed6d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect MC

47e4_0163

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

6037_f28d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

e73e_59f5

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct MC

df82_d356

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect MC

4873_d955

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

8802_4698

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly six (6) boys ♂ and two (2) girls ♀?

⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  8  ⟯
6
(¾)6⋅(¼)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¾)6⋅(¼)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4642 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
6
(¼)6⋅(¾)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¼)6⋅(¾)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 32 
 4642 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
6
(½)6⋅(½)2 = 
 8! 
 (8–6)! ⋅ 6! 
(½)8 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

df82_7359

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect MC

99b6_7163

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

2054_b595

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct MC

b96d_b235

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect MC

99b6_8ea0

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect MC

87d6_8800

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect MC

b96d_d91f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct MC

df77_c31f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect MC

2054_89a3

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

b96d_16b9

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect MC

47e4_496c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

f896_b731

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

df82_05b6

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect MC

6037_81f1

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

87d6_c0c6

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

2ade_305d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect MC

cdd7_53c1

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct MC

2ade_ff3b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect MC

4873_dcd2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect MC

b96d_170a

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect MC

e73e_e3b5

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect MC

4873_a354

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

68e0_715e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

99b6_7576

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect MC

2054_2c87

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect MC

2ade_56f1

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

b96d_e887

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect MC

7c5b_2215

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect MC

99b6_7750

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect MC

6037_f30b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

87d6_9906

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

68e0_197d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

b96d_fea2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

d2c1_306d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct MC

b96d_f628

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct MC

7c5b_a2be

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect MC

99b6_b2bb

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect MC

47e4_450f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect MC

6037_c083

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

e73e_71c3

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct MC

42bf_0a46

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly six (6) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
6
(½)6⋅(½)4 = 
 10! 
 (10–6)! ⋅ 6! 
(½)10 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
6
(¾)6⋅(¼)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¾)6⋅(¼)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4644 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
6
(¼)6⋅(¾)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¼)6⋅(¾)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 34 
 4644 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect MC

7c5b_2bcb

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect MC

d2c1_9985

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct MC

2ade_d090

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

082d_67dd

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly three (3) boys ♂ and five (5) girls ♀?

⟮  8  ⟯
3
(¾)3⋅(¼)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¾)3⋅(¼)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4345 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  8  ⟯
3
(½)3⋅(½)5 = 
 8! 
 (8–3)! ⋅ 3! 
(½)8 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
3
(¼)3⋅(¾)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¼)3⋅(¾)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 35 
 4345 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

3dda_650d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct MC

2ade_be8c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect MC

47e4_2252

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect MC

cdd7_5d11

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect MC

4873_cbff

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect MC

3dda_b011

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect MC

7c5b_7e94

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect MC

8802_d748

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly six (6) boys ♂ and two (2) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
6
(¼)6⋅(¾)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¼)6⋅(¾)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 32 
 4642 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
6
(½)6⋅(½)2 = 
 8! 
 (8–6)! ⋅ 6! 
(½)8 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  8  ⟯
6
(¾)6⋅(¼)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¾)6⋅(¼)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4642 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect MC

df82_b0bb

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct MC

ea1b_9343

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly five (5) boys ♂ and four (4) girls ♀?

⟮  9  ⟯
5
(¼)5⋅(¾)4 = 
 9! 
 (9–5)! ⋅ 5! 
(¼)5⋅(¾)4 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 34 
 4544 
 = 
 126×81 
 262144 
 = 0.0389 = 3.9%
Incorrect
⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect
⟮  9  ⟯
5
(¾)5⋅(¼)4 = 
 9! 
 (9–5)! ⋅ 5! 
(¾)5⋅(¼)4 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4544 
 = 
 126×243 
 262144 
 = 0.1168 = 11.7%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
5
(½)5⋅(½)4 = 
 9! 
 (9–5)! ⋅ 5! 
(½)9 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 126 
 512 
 = 0.2461 = 24.6%
Correct MC

87d6_4da0

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect MC

f896_c30b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

0019_2910

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect MC

47e4_3c89

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct MC

2054_f2dd

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect MC

6037_6790

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect MC

7c5b_a817

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect MC

7c5b_d941

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct MC

df77_960b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect MC

d2c1_fc0e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

99b6_6898

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect MC

0019_63a6

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct MC

68e0_178a

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct MC

2ade_5b59

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

ef84_1ade

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly four (4) boys ♂ and four (4) girls ♀?

⟮  8  ⟯
4
(¾)4⋅(¼)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¾)4⋅(¼)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  8  ⟯
4
(½)4⋅(½)4 = 
 8! 
 (8–4)! ⋅ 4! 
(½)8 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 28 
 = 
 70 
 256 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)4 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
4
(¼)4⋅(¾)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¼)4⋅(¾)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect MC

e73e_3ef1

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

d771_1b53

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect MC

42bf_3f6b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly six (6) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
6
(½)6⋅(½)4 = 
 10! 
 (10–6)! ⋅ 6! 
(½)10 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
6
(¼)6⋅(¾)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¼)6⋅(¾)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 34 
 4644 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
6
(¾)6⋅(¼)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¾)6⋅(¼)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4644 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect MC

691d_da43

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect MC

691d_948a

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect MC

87d6_884b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect MC

df82_07a9

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct MC

e73e_9ebf

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect MC

4873_addd

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect MC

99b6_c83d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

691d_7b48

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct MC

d771_4ff8

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect MC

13f2_5d81

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly five (5) boys ♂ and five (5) girls ♀?

⟮  10  ⟯
5
(½)5⋅(½)5 = 
 10! 
 (10–5)! ⋅ 5! 
(½)10 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 252 
 1024 
 = 0.2461 = 24.6%
Correct
⟮  5  ⟯
5
(½)5⋅(½)5 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
5
(¾)5⋅(¼)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¾)5⋅(¼)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect
⟮  10  ⟯
5
(¼)5⋅(¾)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¼)5⋅(¾)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect MC

2ade_b81f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

b96d_3a65

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

87d6_3708

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

47e4_ca31

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect MC

8802_c68e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly six (6) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
6
(½)6⋅(½)2 = 
 8! 
 (8–6)! ⋅ 6! 
(½)8 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  8  ⟯
6
(¼)6⋅(¾)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¼)6⋅(¾)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 32 
 4642 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
6
(¾)6⋅(¼)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¾)6⋅(¼)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4642 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect MC

df82_9a2f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct MC

691d_fe0b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect MC

b96d_91f2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

4873_6692

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct MC

d771_f324

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect MC

2054_98aa

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect MC

b96d_f786

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

0019_61c2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

0019_0229

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect MC

87d6_df3e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

99b6_a404

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct MC

082d_8f05

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly three (3) boys ♂ and five (5) girls ♀?

⟮  8  ⟯
3
(½)3⋅(½)5 = 
 8! 
 (8–3)! ⋅ 3! 
(½)8 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
3
(¼)3⋅(¾)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¼)3⋅(¾)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 35 
 4345 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
3
(¾)3⋅(¼)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¾)3⋅(¼)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4345 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect MC

082d_0417

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly three (3) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
3
(½)3⋅(½)5 = 
 8! 
 (8–3)! ⋅ 3! 
(½)8 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
3
(¾)3⋅(¼)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¾)3⋅(¼)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4345 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  8  ⟯
3
(¼)3⋅(¾)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¼)3⋅(¾)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 35 
 4345 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

d2c1_d7b9

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect MC

42bf_1491

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly six (6) boys ♂ and four (4) girls ♀?

⟮  10  ⟯
6
(¾)6⋅(¼)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¾)6⋅(¼)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4644 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
6
(½)6⋅(½)4 = 
 10! 
 (10–6)! ⋅ 6! 
(½)10 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
6
(¼)6⋅(¾)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¼)6⋅(¾)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 34 
 4644 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

42bf_b19b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly six (6) boys ♂ and four (4) girls ♀?

⟮  10  ⟯
6
(¾)6⋅(¼)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¾)6⋅(¼)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4644 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  10  ⟯
6
(¼)6⋅(¾)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¼)6⋅(¾)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 34 
 4644 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
6
(½)6⋅(½)4 = 
 10! 
 (10–6)! ⋅ 6! 
(½)10 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct MC

df77_db9d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

d2c1_6278

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect MC

6037_e129

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct MC

cdd7_5c80

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

df82_8f48

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct MC

691d_a28c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect MC

691d_c503

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect MC

d771_f807

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect MC

b96d_3891

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly two (2) boys ♂ and five (5) girls ♀?

⟮  7  ⟯
2
(¼)2⋅(¾)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¼)2⋅(¾)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 35 
 4245 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)5 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
2
(¾)2⋅(¼)5 = 
 7! 
 (7–2)! ⋅ 2! 
(¾)2⋅(¼)5 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4245 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect MC

f896_60d5

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct MC

2054_4254

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect MC

2ade_65c2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

691d_006f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect MC

3dda_bd7c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct MC

df82_4131

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect MC

082d_0a0e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly three (3) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
3
(½)3⋅(½)5 = 
 8! 
 (8–3)! ⋅ 3! 
(½)8 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
3
(¼)3⋅(¾)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¼)3⋅(¾)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 35 
 4345 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
3
(¾)3⋅(¼)5 = 
 8! 
 (8–3)! ⋅ 3! 
(¾)3⋅(¼)5 = 
 8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4345 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect MC

4873_e05f

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect MC

99b6_ffcf

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly four (4) boys ♂ and six (6) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
4
(½)4⋅(½)6 = 
 10! 
 (10–4)! ⋅ 4! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
4
(¾)4⋅(¼)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¾)4⋅(¼)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4446 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  10  ⟯
4
(¼)4⋅(¾)6 = 
 10! 
 (10–4)! ⋅ 4! 
(¼)4⋅(¾)6 = 
 10⋅9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 36 
 4446 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect MC

87d6_fecd

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

87d6_0452

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

4873_15de

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct MC

df82_02de

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly three (3) boys ♂ and three (3) girls ♀?

⟮  6  ⟯
3
(½)3⋅(½)3 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 64 
 = 0.3125 = 31.2%
Correct
⟮  6  ⟯
3
(¾)3⋅(¼)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¾)3⋅(¼)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)3 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3 = 
 6! 
 (6–3)! ⋅ 3! 
(¼)3⋅(¾)3 = 
 6⋅5⋅4 
 3⋅2 
×
 33 
 4343 
 = 
 20×27 
 4096 
 = 0.1318 = 13.2%
Incorrect MC

6037_e164

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

2ade_b363

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

df77_bb6d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect MC

2ade_08eb

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect MC

6037_be13

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly five (5) boys ♂ and three (3) girls ♀?

⟮  8  ⟯
5
(½)5⋅(½)3 = 
 8! 
 (8–5)! ⋅ 5! 
(½)8 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 56 
 256 
 = 0.2188 = 21.9%
Correct
⟮  8  ⟯
5
(¾)5⋅(¼)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¾)5⋅(¼)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4543 
 = 
 56×243 
 65536 
 = 0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(¼)5⋅(¾)3 = 
 8! 
 (8–5)! ⋅ 5! 
(¼)5⋅(¾)3 = 
 8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 33 
 4543 
 = 
 56×27 
 65536 
 = 0.0231 = 2.3%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)3 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)5 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 256 
 = 0.0391 = 3.9%
Incorrect MC

ea1b_0531

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly five (5) boys ♂ and four (4) girls ♀?

⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect
⟮  9  ⟯
5
(¾)5⋅(¼)4 = 
 9! 
 (9–5)! ⋅ 5! 
(¾)5⋅(¼)4 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4544 
 = 
 126×243 
 262144 
 = 0.1168 = 11.7%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
5
(¼)5⋅(¾)4 = 
 9! 
 (9–5)! ⋅ 5! 
(¼)5⋅(¾)4 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 34 
 4544 
 = 
 126×81 
 262144 
 = 0.0389 = 3.9%
Incorrect
⟮  9  ⟯
5
(½)5⋅(½)4 = 
 9! 
 (9–5)! ⋅ 5! 
(½)9 = 
 9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 126 
 512 
 = 0.2461 = 24.6%
Correct MC

2ade_a2b7

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

2ade_c1f6

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

7c5b_0398

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly six (6) boys ♂ and three (3) girls ♀?

⟮  9  ⟯
6
(¼)6⋅(¾)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¼)6⋅(¾)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 33 
 4643 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  9  ⟯
6
(¾)6⋅(¼)3 = 
 9! 
 (9–6)! ⋅ 6! 
(¾)6⋅(¼)3 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4643 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect
⟮  9  ⟯
6
(½)6⋅(½)3 = 
 9! 
 (9–6)! ⋅ 6! 
(½)9 = 
 9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct MC

ef84_d226

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly four (4) boys ♂ and four (4) girls ♀?

⟮  8  ⟯
4
(¼)4⋅(¾)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¼)4⋅(¾)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  8  ⟯
4
(¾)4⋅(¼)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¾)4⋅(¼)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  8  ⟯
4
(½)4⋅(½)4 = 
 8! 
 (8–4)! ⋅ 4! 
(½)8 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 28 
 = 
 70 
 256 
 = 0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(½)4⋅(½)4 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

47e4_5ed4

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly two (2) boys ♂ and eight (8) girls ♀?

⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
2
(¾)2⋅(¼)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¾)2⋅(¼)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4248 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  10  ⟯
2
(¼)2⋅(¾)8 = 
 10! 
 (10–2)! ⋅ 2! 
(¼)2⋅(¾)8 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 38 
 4248 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
2
(½)2⋅(½)8 = 
 10! 
 (10–2)! ⋅ 2! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct MC

5051_1c8a

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly four (4) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect
⟮  9  ⟯
4
(¾)4⋅(¼)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¾)4⋅(¼)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4445 
 = 
 126×81 
 262144 
 = 0.0389 = 3.9%
Incorrect
⟮  9  ⟯
4
(½)4⋅(½)5 = 
 9! 
 (9–4)! ⋅ 4! 
(½)9 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 29 
 = 
 126 
 512 
 = 0.2461 = 24.6%
Correct
⟮  9  ⟯
4
(¼)4⋅(¾)5 = 
 9! 
 (9–4)! ⋅ 4! 
(¼)4⋅(¾)5 = 
 9⋅8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 35 
 4445 
 = 
 126×243 
 262144 
 = 0.1168 = 11.7%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

8802_dcbf

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly six (6) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
6
(¼)6⋅(¾)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¼)6⋅(¾)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 32 
 4642 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect
⟮  8  ⟯
6
(¾)6⋅(¼)2 = 
 8! 
 (8–6)! ⋅ 6! 
(¾)6⋅(¼)2 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4642 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
6
(½)6⋅(½)2 = 
 8! 
 (8–6)! ⋅ 6! 
(½)8 = 
 8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct MC

e73e_c13b

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly seven (7) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
7
(½)7⋅(½)3 = 
 10! 
 (10–7)! ⋅ 7! 
(½)10 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  10  ⟯
7
(¾)7⋅(¼)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¾)7⋅(¼)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4743 
 =  = 0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(¼)7⋅(¾)3 = 
 10! 
 (10–7)! ⋅ 7! 
(¼)7⋅(¾)3 = 
 10⋅9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 33 
 4743 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect MC

13f2_ad1d

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly five (5) boys ♂ and five (5) girls ♀?

⟮  10  ⟯
5
(¼)5⋅(¾)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¼)5⋅(¾)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5 = 
 10! 
 (10–5)! ⋅ 5! 
(½)10 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 252 
 1024 
 = 0.2461 = 24.6%
Correct
⟮  5  ⟯
5
(½)5⋅(½)5 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
5
(¾)5⋅(¼)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¾)5⋅(¼)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect MC

68e0_c6f2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct MC

3dda_3ef3

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect MC

cdd7_8b52

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect MC

87d6_52a2

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

2054_cde7

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly three (3) boys ♂ and six (6) girls ♀?

⟮  9  ⟯
3
(¼)3⋅(¾)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¼)3⋅(¾)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 36 
 4346 
 = 
 84×729 
 262144 
 = 0.2336 = 23.4%
Incorrect
⟮  9  ⟯
3
(½)3⋅(½)6 = 
 9! 
 (9–3)! ⋅ 3! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 29 
 = 
 84 
 512 
 = 0.1641 = 16.4%
Correct
⟮  6  ⟯
6
(½)6⋅(½)3 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
3
(¾)3⋅(¼)6 = 
 9! 
 (9–3)! ⋅ 3! 
(¾)3⋅(¼)6 = 
 9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4346 
 = 
 84×27 
 262144 
 = 0.0087 = 0.9%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)6 = 
 6! 
 (6–3)! ⋅ 3! 
(½)6 = 
 6⋅5⋅4 
 3⋅2 
×
 1 
 26 
 = 
 20 
 512 
 = 0.0391 = 3.9%
Incorrect MC

f896_4797

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly seven (7) boys ♂ and two (2) girls ♀?

⟮  9  ⟯
7
(¾)7⋅(¼)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¾)7⋅(¼)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 37 
 4742 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
7
(½)7⋅(½)2 = 
 9! 
 (9–7)! ⋅ 7! 
(½)9 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect
⟮  9  ⟯
7
(¼)7⋅(¾)2 = 
 9! 
 (9–7)! ⋅ 7! 
(¼)7⋅(¾)2 = 
 9⋅8 
 7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4742 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

691d_f8ad

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly two (2) boys ♂ and four (4) girls ♀?

⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)4 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  6  ⟯
2
(¼)2⋅(¾)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¼)2⋅(¾)4 = 
 6⋅5⋅4⋅3 
 2 
×
 34 
 4244 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  6  ⟯
2
(¾)2⋅(¼)4 = 
 6! 
 (6–2)! ⋅ 2! 
(¾)2⋅(¼)4 = 
 6⋅5⋅4⋅3 
 2 
×
 32 
 4244 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect MC

4873_cf75

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect MC

87d6_0f66

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly three (3) boys ♂ and two (2) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
3
(¼)3⋅(¾)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¼)3⋅(¾)2 = 
 5⋅4 
 3⋅2 
×
 32 
 4342 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  5  ⟯
3
(¾)3⋅(¼)2 = 
 5! 
 (5–3)! ⋅ 3! 
(¾)3⋅(¼)2 = 
 5⋅4 
 3⋅2 
×
 33 
 4342 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(½)3⋅(½)2 = 
 5! 
 (5–3)! ⋅ 3! 
(½)5 = 
 5⋅4 
 3⋅2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

d771_496e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect MC

68e0_93dc

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect MC

68e0_aa02

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly eight (8) boys ♂ and two (2) girls ♀?

⟮  8  ⟯
8
(½)8⋅(½)2 = 
 8! 
 (8–8)! ⋅ 8! 
(½)8 = 
 1 
 1 
×
 1 
 28 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
8
(¾)8⋅(¼)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¾)8⋅(¼)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 38 
 4842 
 =  = 0.2816 = 28.2%
Incorrect
⟮  10  ⟯
8
(¼)8⋅(¾)2 = 
 10! 
 (10–8)! ⋅ 8! 
(¼)8⋅(¾)2 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 32 
 4842 
 = 
 45×9 
 1048576 
 = 0.0004 = 0.0%
Incorrect
⟮  10  ⟯
8
(½)8⋅(½)2 = 
 10! 
 (10–8)! ⋅ 8! 
(½)10 = 
 10⋅9 
 8⋅7⋅6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 45 
 1024 
 = 0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)2⋅(½)8 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 1024 
 = 0.0273 = 2.7%
Incorrect MC

ef84_b8b5

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly four (4) boys ♂ and four (4) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)4 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
4
(½)4⋅(½)4 = 
 8! 
 (8–4)! ⋅ 4! 
(½)8 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 28 
 = 
 70 
 256 
 = 0.2734 = 27.3%
Correct
⟮  8  ⟯
4
(¼)4⋅(¾)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¼)4⋅(¾)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  8  ⟯
4
(¾)4⋅(¼)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¾)4⋅(¼)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect MC

2ade_3964

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has five (5) children. What is the probability that she has exactly two (2) boys ♂ and three (3) girls ♀?

⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct
⟮  5  ⟯
2
(¼)2⋅(¾)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¼)2⋅(¾)3 = 
 5⋅4⋅3 
 2 
×
 33 
 4243 
 = 
 10×27 
 1024 
 = 0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(¾)2⋅(¼)3 = 
 5! 
 (5–2)! ⋅ 2! 
(¾)2⋅(¼)3 = 
 5⋅4⋅3 
 2 
×
 32 
 4243 
 = 
 10×9 
 1024 
 = 0.0879 = 8.8%
Incorrect
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

ef84_6fda

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly four (4) boys ♂ and four (4) girls ♀?

⟮  8  ⟯
4
(¼)4⋅(¾)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¼)4⋅(¾)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  8  ⟯
4
(¾)4⋅(¼)4 = 
 8! 
 (8–4)! ⋅ 4! 
(¾)4⋅(¼)4 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4444 
 = 
 70×81 
 65536 
 = 0.0865 = 8.7%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)4 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
4
(½)4⋅(½)4 = 
 8! 
 (8–4)! ⋅ 4! 
(½)8 = 
 8⋅7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 28 
 = 
 70 
 256 
 = 0.2734 = 27.3%
Correct MC

0019_c5c6

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly three (3) boys ♂ and four (4) girls ♀?

⟮  7  ⟯
3
(½)3⋅(½)4 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct
⟮  7  ⟯
3
(¼)3⋅(¾)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¼)3⋅(¾)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 34 
 4344 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  7  ⟯
3
(¾)3⋅(¼)4 = 
 7! 
 (7–3)! ⋅ 3! 
(¾)3⋅(¼)4 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4344 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect MC

4873_4257

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has nine (9) children. What is the probability that she has exactly two (2) boys ♂ and seven (7) girls ♀?

⟮  9  ⟯
2
(¾)2⋅(¼)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¾)2⋅(¼)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4247 
 = 
 36×9 
 262144 
 = 0.0012 = 0.1%
Incorrect
⟮  9  ⟯
2
(¼)2⋅(¾)7 = 
 9! 
 (9–2)! ⋅ 2! 
(¼)2⋅(¾)7 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 37 
 4247 
 =  = 0.3003 = 30.0%
Incorrect
⟮  9  ⟯
2
(½)2⋅(½)7 = 
 9! 
 (9–2)! ⋅ 2! 
(½)9 = 
 9⋅8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 29 
 = 
 36 
 512 
 = 0.0703 = 7.0%
Correct
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect
⟮  7  ⟯
2
(½)2⋅(½)7 = 
 7! 
 (7–2)! ⋅ 2! 
(½)7 = 
 7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 27 
 = 
 21 
 512 
 = 0.0410 = 4.1%
Incorrect MC

d771_6a9c

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has six (6) children. What is the probability that she has exactly four (4) boys ♂ and two (2) girls ♀?

⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect
⟮  6  ⟯
4
(¾)4⋅(¼)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¾)4⋅(¼)2 = 
 6⋅5 
 4⋅3⋅2 
×
 34 
 4442 
 = 
 15×81 
 4096 
 = 0.2966 = 29.7%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)4 = 
 4! 
 (4–2)! ⋅ 2! 
(½)4 = 
 4⋅3 
 2 
×
 1 
 24 
 = 
 6 
 64 
 = 0.0938 = 9.4%
Incorrect
⟮  6  ⟯
4
(¼)4⋅(¾)2 = 
 6! 
 (6–4)! ⋅ 4! 
(¼)4⋅(¾)2 = 
 6⋅5 
 4⋅3⋅2 
×
 32 
 4442 
 = 
 15×9 
 4096 
 = 0.0330 = 3.3%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)2 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 64 
 = 0.2344 = 23.4%
Correct MC

d2c1_ae55

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly four (4) boys ♂ and three (3) girls ♀?

⟮  7  ⟯
4
(¼)4⋅(¾)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¼)4⋅(¾)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 33 
 4443 
 = 
 35×27 
 16384 
 = 0.0577 = 5.8%
Incorrect
⟮  4  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect
⟮  7  ⟯
4
(¾)4⋅(¼)3 = 
 7! 
 (7–4)! ⋅ 4! 
(¾)4⋅(¼)3 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 34 
 4443 
 = 
 35×81 
 16384 
 = 0.1730 = 17.3%
Incorrect
⟮  4  ⟯
4
(½)4⋅(½)3 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
4
(½)4⋅(½)3 = 
 7! 
 (7–4)! ⋅ 4! 
(½)7 = 
 7⋅6⋅5 
 4⋅3⋅2 
×
 1 
 27 
 = 
 35 
 128 
 = 0.2734 = 27.3%
Correct MC

13f2_6b00

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly five (5) boys ♂ and five (5) girls ♀?

⟮  10  ⟯
5
(¼)5⋅(¾)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¼)5⋅(¾)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect
⟮  5  ⟯
5
(½)5⋅(½)5 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5 = 
 10! 
 (10–5)! ⋅ 5! 
(½)10 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 252 
 1024 
 = 0.2461 = 24.6%
Correct
⟮  10  ⟯
5
(¾)5⋅(¼)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¾)5⋅(¼)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect MC

13f2_c8f0

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly five (5) boys ♂ and five (5) girls ♀?

⟮  5  ⟯
5
(½)5⋅(½)5 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
5
(¾)5⋅(¼)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¾)5⋅(¼)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5 = 
 10! 
 (10–5)! ⋅ 5! 
(½)10 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 252 
 1024 
 = 0.2461 = 24.6%
Correct
⟮  10  ⟯
5
(¼)5⋅(¾)5 = 
 10! 
 (10–5)! ⋅ 5! 
(¼)5⋅(¾)5 = 
 10⋅9⋅8⋅7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4545 
 = 
 252×243 
 1048576 
 = 0.0584 = 5.8%
Incorrect MC

42bf_a8ac

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly six (6) boys ♂ and four (4) girls ♀?

⟮  10  ⟯
6
(¾)6⋅(¼)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¾)6⋅(¼)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 36 
 4644 
 = 
 210×729 
 1048576 
 = 0.1460 = 14.6%
Incorrect
⟮  10  ⟯
6
(½)6⋅(½)4 = 
 10! 
 (10–6)! ⋅ 6! 
(½)10 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 1 
 210 
 = 
 210 
 1024 
 = 0.2051 = 20.5%
Correct
⟮  10  ⟯
6
(¼)6⋅(¾)4 = 
 10! 
 (10–6)! ⋅ 6! 
(¼)6⋅(¾)4 = 
 10⋅9⋅8⋅7 
 6⋅5⋅4⋅3⋅2 
×
 34 
 4644 
 = 
 210×81 
 1048576 
 = 0.0162 = 1.6%
Incorrect
⟮  6  ⟯
4
(½)4⋅(½)6 = 
 6! 
 (6–4)! ⋅ 4! 
(½)6 = 
 6⋅5 
 4⋅3⋅2 
×
 1 
 26 
 = 
 15 
 1024 
 = 0.0146 = 1.5%
Incorrect
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

cdd7_8065

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has eight (8) children. What is the probability that she has exactly two (2) boys ♂ and six (6) girls ♀?

⟮  8  ⟯
2
(½)2⋅(½)6 = 
 8! 
 (8–2)! ⋅ 2! 
(½)8 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 1 
 28 
 = 
 28 
 256 
 = 0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect
⟮  8  ⟯
2
(¼)2⋅(¾)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¼)2⋅(¾)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 36 
 4246 
 = 
 28×729 
 65536 
 = 0.3115 = 31.1%
Incorrect
⟮  8  ⟯
2
(¾)2⋅(¼)6 = 
 8! 
 (8–2)! ⋅ 2! 
(¾)2⋅(¼)6 = 
 8⋅7⋅6⋅5⋅4⋅3 
 2 
×
 32 
 4246 
 = 
 28×9 
 65536 
 = 0.0038 = 0.4%
Incorrect
⟮  6  ⟯
2
(½)2⋅(½)6 = 
 6! 
 (6–2)! ⋅ 2! 
(½)6 = 
 6⋅5⋅4⋅3 
 2 
×
 1 
 26 
 = 
 15 
 256 
 = 0.0586 = 5.9%
Incorrect MC

3dda_5190

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has seven (7) children. What is the probability that she has exactly five (5) boys ♂ and two (2) girls ♀?

⟮  7  ⟯
5
(¾)5⋅(¼)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¾)5⋅(¼)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 35 
 4542 
 = 
 21×243 
 16384 
 = 0.3115 = 31.1%
Incorrect
⟮  7  ⟯
5
(½)5⋅(½)2 = 
 7! 
 (7–5)! ⋅ 5! 
(½)7 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 1 
 27 
 = 
 21 
 128 
 = 0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect
⟮  7  ⟯
5
(¼)5⋅(¾)2 = 
 7! 
 (7–5)! ⋅ 5! 
(¼)5⋅(¾)2 = 
 7⋅6 
 5⋅4⋅3⋅2 
×
 32 
 4542 
 = 
 21×9 
 16384 
 = 0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)2⋅(½)5 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 128 
 = 0.0781 = 7.8%
Incorrect MC

df77_8a1e

Model: Binomial →
⟮  n  ⟯
k
⋅pk⋅qn-k

In this scenario, assume that each child is born independently with the same chance of being either sex. The event outcomes are mutually exclusive, so we can apply the binomial model to determine the probability of a specific combination.

A woman has ten (10) children. What is the probability that she has exactly three (3) boys ♂ and seven (7) girls ♀?

⟮  10  ⟯
3
(¾)3⋅(¼)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¾)3⋅(¼)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 33 
 4347 
 = 
 120×27 
 1048576 
 = 0.0031 = 0.3%
Incorrect
⟮  10  ⟯
3
(½)3⋅(½)7 = 
 10! 
 (10–3)! ⋅ 3! 
(½)10 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 210 
 = 
 120 
 1024 
 = 0.1172 = 11.7%
Correct
⟮  7  ⟯
3
(½)3⋅(½)7 = 
 7! 
 (7–3)! ⋅ 3! 
(½)7 = 
 7⋅6⋅5⋅4 
 3⋅2 
×
 1 
 27 
 = 
 35 
 1024 
 = 0.0342 = 3.4%
Incorrect
⟮  7  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect
⟮  10  ⟯
3
(¼)3⋅(¾)7 = 
 10! 
 (10–3)! ⋅ 3! 
(¼)3⋅(¾)7 = 
 10⋅9⋅8⋅7⋅6⋅5⋅4 
 3⋅2 
×
 37 
 4347 
 =  = 0.2503 = 25.0%
Incorrect