MC

df77_b9ca

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
⟮  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
⟮  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
(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

d771_df52

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
⟮  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
⟮  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 MC

42bf_5b5f

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
(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
⟮  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 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

d2c1_000b

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  ⟯
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 MC

df82_66d3

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
(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 MC

2ade_17f4

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
(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
(½)2⋅(½)3 = 
 5! 
 (5–2)! ⋅ 2! 
(½)5 = 
 5⋅4⋅3 
 2 
×
 1 
 25 
 = 
 10 
 32 
 = 0.3125 = 31.2%
Correct MC

87d6_507e

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
(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
⟮  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 MC

d771_6377

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
(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
(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 MC

b96d_f5d4

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  ⟯
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
(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

b96d_b469

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  ⟯
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
⟮  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 MC

e73e_45e6

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
(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
⟮  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 MC

b96d_dee2

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
(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
⟮  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

082d_878a

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
⟮  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
⟮  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 MC

6037_3e11

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
⟮  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
(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

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

ef84_7665

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
(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
(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

ef84_3c4a

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
(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
(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
(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

3dda_de80

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  ⟯
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
⟮  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
⟮  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

2054_da28

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
⟮  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
(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

2ade_05c7

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
(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
⟮  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

b96d_9565

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
(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
⟮  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_b052

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
⟮  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
(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 MC

df82_8d01

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
⟮  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 MC

691d_9eae

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
(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
⟮  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

2ade_f1db

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
⟮  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
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

0019_8e1b

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  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
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
⟮  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 MC

87d6_c912

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
(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
⟮  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_a0ed

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
⟮  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
⟮  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 MC

082d_8d08

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
(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
⟮  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 MC

42bf_7129

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
(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
⟮  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
⟮  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_16b2

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
⟮  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
⟮  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 MC

b96d_c5d1

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
(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 MC

87d6_8d6f

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
(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
⟮  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 MC

4873_d216

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  ⟯
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
(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 MC

b96d_bbf6

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
(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
⟮  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 MC

691d_b5da

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
(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
⟮  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

68e0_761d

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
⟮  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
(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 MC

4873_a22e

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
⟮  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
⟮  7  ⟯
7
(½)7⋅(½)2 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

87d6_85d6

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  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
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

87d6_f4a8

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
(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
⟮  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

b96d_ab29

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
⟮  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  ⟯
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

691d_8eeb

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
⟮  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
⟮  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

cdd7_f43c

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  ⟯
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  ⟯
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 MC

0019_f1ce

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
(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  ⟯
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
⟮  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

082d_0f9d

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
(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
⟮  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

5051_8d39

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
(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
⟮  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 MC

3dda_880d

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
(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
(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
⟮  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

87d6_b4c9

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
(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
⟮  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 MC

691d_0ad1

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
⟮  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
(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
⟮  4  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
Incorrect MC

d771_7d5f

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
⟮  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
⟮  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

2ade_d8a2

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
(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  ⟯
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 MC

7c5b_6727

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
(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
⟮  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

0019_be3a

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
(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
⟮  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 MC

2054_27e6

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
(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
⟮  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 MC

13f2_d0be

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
(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
⟮  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
(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

cdd7_4e15

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  ⟯
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  ⟯
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

68e0_426d

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
⟮  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
⟮  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

5051_f1fe

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  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
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
(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
⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect MC

7c5b_cab5

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
⟮  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
(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

87d6_ba98

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
(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
⟮  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 MC

b96d_3afc

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
(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  ⟯
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

691d_eaae

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  ⟯
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
⟮  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

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

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_f076

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
⟮  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 MC

df77_7e0e

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
⟮  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
(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
(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 MC

87d6_729f

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
(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

5051_8844

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
(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
⟮  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

7c5b_56cd

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  ⟯
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
(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
⟮  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

87d6_4685

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
(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
⟮  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_95fa

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
(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
⟮  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
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

47e4_e0a0

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
(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
(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 MC

6037_6e0c

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
⟮  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

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_49e5

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
(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
⟮  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

4873_9ac7

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  ⟯
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
⟮  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

5051_556f

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
(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
⟮  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
⟮  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
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

87d6_2477

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
(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  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

d771_7eaa

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
⟮  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 MC

df82_e4ae

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
⟮  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 MC

8802_0457

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
(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
⟮  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
(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

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

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

082d_560e

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
⟮  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
⟮  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

e73e_5fd8

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
⟮  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
⟮  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

082d_4725

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
⟮  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
⟮  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 MC

2ade_59bc

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
⟮  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
⟮  3  ⟯
2
(½)2⋅(½)3 = 
 3! 
 (3–2)! ⋅ 2! 
(½)3 = 
 3 
 1 
×
 1 
 23 
 = 
 3 
 32 
 = 0.0938 = 9.4%
Incorrect MC

d771_1073

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
⟮  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  ⟯
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 MC

0019_9e52

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  ⟯
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
⟮  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 MC

4873_1ab7

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
⟮  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

cdd7_9375

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
⟮  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
⟮  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

2ade_6fa1

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
⟮  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 MC

3dda_e57e

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
⟮  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
⟮  5  ⟯
5
(½)5⋅(½)2 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 128 
 = 0.0078 = 0.8%
Incorrect MC

cdd7_bcc0

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
(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
⟮  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

d2c1_9546

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
(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
(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 MC

ef84_e905

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
⟮  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
(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 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

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

f896_9667

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
(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
⟮  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
⟮  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

d771_5c2b

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
(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
(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  ⟯
4
(½)4⋅(½)2 = 
 4! 
 (4–4)! ⋅ 4! 
(½)4 = 
 1 
 1 
×
 1 
 24 
 = 
 1 
 64 
 = 0.0156 = 1.6%
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 MC

d771_3593

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
⟮  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
⟮  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 MC

6037_8e73

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  ⟯
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
⟮  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

87d6_0acc

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
(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 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

ef84_1ecb

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
(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
(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 MC

3dda_e04b

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
⟮  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
(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
(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

5051_0786

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
⟮  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
⟮  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
(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

2ade_a501

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
(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
⟮  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

082d_b699

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
(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
⟮  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
(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 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

f896_b36f

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
⟮  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
(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

6037_f031

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  ⟯
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 MC

3dda_aa13

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
⟮  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
⟮  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

d771_83e0

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
⟮  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
⟮  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

d771_91e0

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
(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
(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
(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

f896_2082

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
⟮  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
(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

691d_2aa8

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
(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
⟮  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

b96d_9452

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
(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
⟮  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

47e4_23ac

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
(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
⟮  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

87d6_212d

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
(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
⟮  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
⟮  3  ⟯
3
(½)3⋅(½)2 = 
 3! 
 (3–3)! ⋅ 3! 
(½)3 = 
 1 
 1 
×
 1 
 23 
 = 
 1 
 32 
 = 0.0312 = 3.1%
Incorrect MC

87d6_233d

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
(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
⟮  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
⟮  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 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

082d_2cc3

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
(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
⟮  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

082d_40f5

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
(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
⟮  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 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

082d_8834

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
⟮  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 MC

6037_6847

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
(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
(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
(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

0019_d78d

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
(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
⟮  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 MC

d2c1_6591

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
⟮  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
⟮  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

df77_f0bf

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  ⟯
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
⟮  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
(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  ⟯
7
(½)7⋅(½)3 = 
 7! 
 (7–7)! ⋅ 7! 
(½)7 = 
 1 
 1 
×
 1 
 27 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

d771_7f97

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
⟮  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
⟮  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

f896_26f0

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
⟮  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
(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

691d_d421

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
⟮  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
⟮  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

b96d_0be6

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
⟮  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
(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

87d6_7ef0

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
⟮  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 MC

5051_2ec0

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
⟮  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
(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
⟮  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 MC

cdd7_086f

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  ⟯
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  ⟯
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  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

87d6_9149

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
(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
⟮  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 MC

691d_be1b

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
(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  ⟯
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
(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

ea1b_175e

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
(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
⟮  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  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
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
⟮  5  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
Incorrect MC

4873_e2c4

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
(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 MC

87d6_278e

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
(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
⟮  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 MC

5051_a399

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
⟮  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
⟮  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
⟮  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 MC

5051_410f

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
(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
⟮  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
(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
⟮  5  ⟯
4
(½)4⋅(½)5 = 
 5! 
 (5–4)! ⋅ 4! 
(½)5 = 
 5 
 1 
×
 1 
 25 
 = 
 5 
 512 
 = 0.0098 = 1.0%
Incorrect MC

13f2_798e

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
(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
⟮  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 MC

b96d_bad6

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
⟮  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
⟮  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 MC

0019_9687

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
(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
⟮  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

cdd7_ac4f

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  ⟯
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  ⟯
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  ⟯
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 MC

68e0_74c2

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
(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 MC

df82_9866

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
⟮  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 MC

7c5b_8420

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
⟮  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

b96d_7756

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  ⟯
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
⟮  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

42bf_3b03

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
(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
⟮  6  ⟯
6
(½)6⋅(½)4 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
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
⟮  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

6037_902a

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
⟮  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
⟮  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

f896_4d86

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
(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
⟮  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
⟮  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_ed82

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
(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  ⟯
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
⟮  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

2054_0013

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
⟮  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
(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  ⟯
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

4873_db75

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
⟮  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
⟮  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

68e0_3cf4

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
(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
⟮  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
⟮  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 MC

b96d_9e5d

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
(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 MC

d2c1_3f01

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
⟮  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
⟮  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 MC

d2c1_32d8

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  ⟯
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
⟮  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 MC

691d_a845

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
(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
⟮  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
⟮  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

13f2_5153

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
⟮  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
(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
⟮  5  ⟯
5
(½)5⋅(½)5 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 1024 
 = 0.0010 = 0.1%
Incorrect MC

d771_3630

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
⟮  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
⟮  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
(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 MC

0019_ed0a

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
(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  ⟯
3
(½)3⋅(½)4 = 
 4! 
 (4–3)! ⋅ 3! 
(½)4 = 
 4 
 1 
×
 1 
 24 
 = 
 4 
 128 
 = 0.0312 = 3.1%
Incorrect MC

8802_bec6

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
(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
⟮  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
(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

0019_b376

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  ⟯
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
(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 MC

5051_feb9

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
⟮  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
⟮  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
(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

d771_7a2a

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
⟮  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 MC

5051_de63

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  ⟯
5
(½)5⋅(½)4 = 
 5! 
 (5–5)! ⋅ 5! 
(½)5 = 
 1 
 1 
×
 1 
 25 
 = 
 1 
 512 
 = 0.0020 = 0.2%
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  ⟯
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 MC

f896_9bbd

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
⟮  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
⟮  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  ⟯
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_393a

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  ⟯
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

8802_6bf8

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
(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  ⟯
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
⟮  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

f896_3c20

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
(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  ⟯
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

b96d_90dd

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
⟮  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
(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 MC

cdd7_019a

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
⟮  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  ⟯
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
⟮  6  ⟯
6
(½)6⋅(½)2 = 
 6! 
 (6–6)! ⋅ 6! 
(½)6 = 
 1 
 1 
×
 1 
 26 
 = 
 1 
 256 
 = 0.0039 = 0.4%
Incorrect MC

df77_5a57

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
⟮  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
(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 MC

3dda_2044

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
(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
(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
⟮  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 MC

87d6_4b3f

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
⟮  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
(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

df77_d44a

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
⟮  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 MC

d771_7d2b

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
⟮  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
(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
⟮  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

6037_cac1

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  ⟯
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  ⟯
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
(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
(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

2ade_73f0

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
⟮  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

d2c1_b725

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
⟮  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 MC

68e0_a07f

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
⟮  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

87d6_22a3

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
(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  ⟯
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

87d6_3ad2

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
(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
⟮  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

b96d_6978

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
⟮  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 MC

8802_37f1

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
⟮  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
⟮  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

3dda_06d5

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
(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
(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
(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

42bf_5a51

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
⟮  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
⟮  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

df82_a3fd

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
(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 MC

d771_d080

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
⟮  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
⟮  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 MC

cdd7_0acd

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  ⟯
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  ⟯
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 MC

2ade_15d8

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
⟮  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 MC

cdd7_c259

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
⟮  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  ⟯
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 MC

082d_0c57

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
(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
⟮  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
(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