MC

0f34_c758

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
7
(¼)0⋅(¾)7=
7!
(7–7)! ⋅ 7!
(¼)0⋅(¾)7=
1
1
×
10 ⋅ 37
40 ⋅ 47
=
1 × 2187
16384
=0.1335 = 13.3%
Incorrect
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect
⟮  7  ⟯
4
(½)3⋅(½)4=
7!
(7–4)! ⋅ 4!
(½)3⋅(½)4=
7⋅6⋅5
3⋅2
×
1
27
=
35
128
=0.2734 = 27.3%
Correct
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  4  ⟯
4
(½)0⋅(½)4=
4!
(4–4)! ⋅ 4!
(½)0⋅(½)4=
1
1
×
1
24
=
1
16
=0.0625 = 6.2%
Incorrect MC

6684_2c75

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  10  ⟯
8
(¼)2⋅(¾)8=
10!
(10–8)! ⋅ 8!
(¼)2⋅(¾)8=
10⋅9
2
×
12 ⋅ 38
42 ⋅ 48
=
45 × 6561
1048576
=0.2816 = 28.2%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect
⟮  10  ⟯
2
(½)8⋅(½)2=
10!
(10–2)! ⋅ 2!
(½)8⋅(½)2=
10⋅9
2
×
1
210
=
45
1024
=0.0439 = 4.4%
Correct MC

e47f_5ce4

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
9
(¼)0⋅(¾)9=
9!
(9–9)! ⋅ 9!
(¼)0⋅(¾)9=
1
1
×
10 ⋅ 39
40 ⋅ 49
=
1 × 19683
262144
=0.0751 = 7.5%
Incorrect
⟮  9  ⟯
5
(¼)4⋅(¾)5=
9!
(9–5)! ⋅ 5!
(¼)4⋅(¾)5=
9⋅8⋅7⋅6
4⋅3⋅2
×
14 ⋅ 35
44 ⋅ 45
=
126 × 243
262144
=0.1168 = 11.7%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  5  ⟯
4
(¼)1⋅(¾)4=
5!
(5–4)! ⋅ 4!
(¼)1⋅(¾)4=
5
1
×
11 ⋅ 34
41 ⋅ 44
=
5 × 81
1024
=0.3955 = 39.6%
Incorrect
⟮  9  ⟯
4
(½)5⋅(½)4=
9!
(9–4)! ⋅ 4!
(½)5⋅(½)4=
9⋅8⋅7⋅6
4⋅3⋅2
×
1
29
=
126
512
=0.2461 = 24.6%
Correct MC

b709_d2ac

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  5  ⟯
2
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  7  ⟯
5
(½)2⋅(½)5=
7!
(7–5)! ⋅ 5!
(½)2⋅(½)5=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  7  ⟯
7
(¼)0⋅(¾)7=
7!
(7–7)! ⋅ 7!
(¼)0⋅(¾)7=
1
1
×
10 ⋅ 37
40 ⋅ 47
=
1 × 2187
16384
=0.1335 = 13.3%
Incorrect
⟮  7  ⟯
2
(¾)5⋅(¼)2=
7!
(7–2)! ⋅ 2!
(¾)5⋅(¼)2=
7⋅6
2
×
35 ⋅ 12
45 ⋅ 42
=
21 × 243
16384
=0.3115 = 31.1%
Incorrect MC

0700_71d2

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)5⋅(¼)3=
8!
(8–3)! ⋅ 3!
(¾)5⋅(¼)3=
8⋅7⋅6
3⋅2
×
35 ⋅ 13
45 ⋅ 43
=
56 × 243
65536
=0.2076 = 20.8%
Incorrect
⟮  8  ⟯
5
(½)3⋅(½)5=
8!
(8–5)! ⋅ 5!
(½)3⋅(½)5=
8⋅7⋅6
3⋅2
×
1
28
=
56
256
=0.2188 = 21.9%
Correct
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  8  ⟯
5
(¾)3⋅(¼)5=
8!
(8–5)! ⋅ 5!
(¾)3⋅(¼)5=
8⋅7⋅6
3⋅2
×
33 ⋅ 15
43 ⋅ 45
=
56 × 27
65536
=0.0231 = 2.3%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect MC

89b2_9144

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  3  ⟯
2
(¾)1⋅(¼)2=
3!
(3–2)! ⋅ 2!
(¾)1⋅(¼)2=
3
1
×
31 ⋅ 12
41 ⋅ 42
=
3 × 3
64
=0.1406 = 14.1%
Incorrect
⟮  3  ⟯
2
(½)1⋅(½)2=
3!
(3–2)! ⋅ 2!
(½)1⋅(½)2=
3
1
×
1
23
=
3
8
=0.3750 = 37.5%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct MC

4980_54bc

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  10  ⟯
8
(½)2⋅(½)8=
10!
(10–8)! ⋅ 8!
(½)2⋅(½)8=
10⋅9
2
×
1
210
=
45
1024
=0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)6⋅(½)2=
8!
(8–2)! ⋅ 2!
(½)6⋅(½)2=
8⋅7
2
×
1
28
=
28
256
=0.1094 = 10.9%
Incorrect
⟮  8  ⟯
2
(¾)6⋅(¼)2=
8!
(8–2)! ⋅ 2!
(¾)6⋅(¼)2=
8⋅7
2
×
36 ⋅ 12
46 ⋅ 42
=
28 × 729
65536
=0.3115 = 31.1%
Incorrect
⟮  8  ⟯
8
(¼)0⋅(¾)8=
8!
(8–8)! ⋅ 8!
(¼)0⋅(¾)8=
1
1
×
10 ⋅ 38
40 ⋅ 48
=
1 × 6561
65536
=0.1001 = 10.0%
Incorrect MC

1d96_1315

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  10  ⟯
5
(¼)5⋅(¾)5=
10!
(10–5)! ⋅ 5!
(¼)5⋅(¾)5=
10⋅9⋅8⋅7⋅6
5⋅4⋅3⋅2
×
15 ⋅ 35
45 ⋅ 45
=
252 × 243
1048576
=0.0584 = 5.8%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5=
10!
(10–5)! ⋅ 5!
(½)5⋅(½)5=
10⋅9⋅8⋅7⋅6
5⋅4⋅3⋅2
×
1
210
=
252
1024
=0.2461 = 24.6%
Correct
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect MC

6684_9dab

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  10  ⟯
8
(¼)2⋅(¾)8=
10!
(10–8)! ⋅ 8!
(¼)2⋅(¾)8=
10⋅9
2
×
12 ⋅ 38
42 ⋅ 48
=
45 × 6561
1048576
=0.2816 = 28.2%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect
⟮  10  ⟯
2
(½)8⋅(½)2=
10!
(10–2)! ⋅ 2!
(½)8⋅(½)2=
10⋅9
2
×
1
210
=
45
1024
=0.0439 = 4.4%
Correct
⟮  8  ⟯
2
(½)6⋅(½)2=
8!
(8–2)! ⋅ 2!
(½)6⋅(½)2=
8⋅7
2
×
1
28
=
28
256
=0.1094 = 10.9%
Incorrect
⟮  8  ⟯
8
(¼)0⋅(¾)8=
8!
(8–8)! ⋅ 8!
(¼)0⋅(¾)8=
1
1
×
10 ⋅ 38
40 ⋅ 48
=
1 × 6561
65536
=0.1001 = 10.0%
Incorrect MC

4617_5cef

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
7
(¼)2⋅(¾)7=
9!
(9–7)! ⋅ 7!
(¼)2⋅(¾)7=
9⋅8
2
×
12 ⋅ 37
42 ⋅ 47
=
36 × 2187
262144
=0.3003 = 30.0%
Incorrect
⟮  9  ⟯
9
(¼)0⋅(¾)9=
9!
(9–9)! ⋅ 9!
(¼)0⋅(¾)9=
1
1
×
10 ⋅ 39
40 ⋅ 49
=
1 × 19683
262144
=0.0751 = 7.5%
Incorrect
⟮  9  ⟯
7
(½)2⋅(½)7=
9!
(9–7)! ⋅ 7!
(½)2⋅(½)7=
9⋅8
2
×
1
29
=
36
512
=0.0703 = 7.0%
Correct
⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  7  ⟯
2
(½)5⋅(½)2=
7!
(7–2)! ⋅ 2!
(½)5⋅(½)2=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Incorrect MC

b709_f03b

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  7  ⟯
5
(½)2⋅(½)5=
7!
(7–5)! ⋅ 5!
(½)2⋅(½)5=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  7  ⟯
7
(¼)0⋅(¾)7=
7!
(7–7)! ⋅ 7!
(¼)0⋅(¾)7=
1
1
×
10 ⋅ 37
40 ⋅ 47
=
1 × 2187
16384
=0.1335 = 13.3%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  5  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect MC

2e30_796a

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
6
(½)3⋅(½)6=
9!
(9–6)! ⋅ 6!
(½)3⋅(½)6=
9⋅8⋅7
3⋅2
×
1
29
=
84
512
=0.1641 = 16.4%
Correct
⟮  9  ⟯
6
(¼)3⋅(¾)6=
9!
(9–6)! ⋅ 6!
(¼)3⋅(¾)6=
9⋅8⋅7
3⋅2
×
13 ⋅ 36
43 ⋅ 46
=
84 × 729
262144
=0.2336 = 23.4%
Incorrect
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
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 ⋅ 13
43 ⋅ 43
=
20 × 27
4096
=0.1318 = 13.2%
Incorrect MC

6a4a_330f

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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
(½)4⋅(½)4=
8!
(8–4)! ⋅ 4!
(½)4⋅(½)4=
8⋅7⋅6⋅5
4⋅3⋅2
×
1
28
=
70
256
=0.2734 = 27.3%
Correct
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect
⟮  4  ⟯
4
(½)0⋅(½)4=
4!
(4–4)! ⋅ 4!
(½)0⋅(½)4=
1
1
×
1
24
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  8  ⟯
8
(¼)0⋅(¾)8=
8!
(8–8)! ⋅ 8!
(¼)0⋅(¾)8=
1
1
×
10 ⋅ 38
40 ⋅ 48
=
1 × 6561
65536
=0.1001 = 10.0%
Incorrect
⟮  8  ⟯
4
(¾)4⋅(¼)4=
8!
(8–4)! ⋅ 4!
(¾)4⋅(¼)4=
8⋅7⋅6⋅5
4⋅3⋅2
×
34 ⋅ 14
44 ⋅ 44
=
70 × 81
65536
=0.0865 = 8.7%
Incorrect MC

89b2_09a9

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect MC

7ad5_4efd

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)4⋅(¼)2=
6!
(6–2)! ⋅ 2!
(¾)4⋅(¼)2=
6⋅5
2
×
34 ⋅ 12
44 ⋅ 42
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  6  ⟯
4
(½)2⋅(½)4=
6!
(6–4)! ⋅ 4!
(½)2⋅(½)4=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Correct
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  6  ⟯
4
(¾)2⋅(¼)4=
6!
(6–4)! ⋅ 4!
(¾)2⋅(¼)4=
6⋅5
2
×
32 ⋅ 14
42 ⋅ 44
=
15 × 9
4096
=0.0330 = 3.3%
Incorrect
⟮  4  ⟯
2
(½)2⋅(½)2=
4!
(4–2)! ⋅ 2!
(½)2⋅(½)2=
4⋅3
2
×
1
24
=
6
16
=0.3750 = 37.5%
Incorrect MC

5a4c_498a

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3=
6!
(6–3)! ⋅ 3!
(½)3⋅(½)3=
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 ⋅ 13
43 ⋅ 43
=
20 × 27
4096
=0.1318 = 13.2%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect MC

7ad5_45ad

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
4
(½)2⋅(½)4=
6!
(6–4)! ⋅ 4!
(½)2⋅(½)4=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Correct
⟮  4  ⟯
2
(½)2⋅(½)2=
4!
(4–2)! ⋅ 2!
(½)2⋅(½)2=
4⋅3
2
×
1
24
=
6
16
=0.3750 = 37.5%
Incorrect
⟮  6  ⟯
2
(¾)4⋅(¼)2=
6!
(6–2)! ⋅ 2!
(¾)4⋅(¼)2=
6⋅5
2
×
34 ⋅ 12
44 ⋅ 42
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect MC

0f34_30f2

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)1⋅(½)3=
4!
(4–3)! ⋅ 3!
(½)1⋅(½)3=
4
1
×
1
24
=
4
16
=0.2500 = 25.0%
Incorrect
⟮  7  ⟯
4
(¼)3⋅(¾)4=
7!
(7–4)! ⋅ 4!
(¼)3⋅(¾)4=
7⋅6⋅5
3⋅2
×
13 ⋅ 34
43 ⋅ 44
=
35 × 81
16384
=0.1730 = 17.3%
Incorrect
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  7  ⟯
7
(¼)0⋅(¾)7=
7!
(7–7)! ⋅ 7!
(¼)0⋅(¾)7=
1
1
×
10 ⋅ 37
40 ⋅ 47
=
1 × 2187
16384
=0.1335 = 13.3%
Incorrect
⟮  7  ⟯
4
(½)3⋅(½)4=
7!
(7–4)! ⋅ 4!
(½)3⋅(½)4=
7⋅6⋅5
3⋅2
×
1
27
=
35
128
=0.2734 = 27.3%
Correct MC

89b2_1be6

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect
⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct
⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect MC

33be_c59d

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  5  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect
⟮  7  ⟯
2
(½)5⋅(½)2=
7!
(7–2)! ⋅ 2!
(½)5⋅(½)2=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  7  ⟯
5
(¼)2⋅(¾)5=
7!
(7–5)! ⋅ 5!
(¼)2⋅(¾)5=
7⋅6
2
×
12 ⋅ 35
42 ⋅ 45
=
21 × 243
16384
=0.3115 = 31.1%
Incorrect MC

7ad5_97c6

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
4
(½)2⋅(½)4=
6!
(6–4)! ⋅ 4!
(½)2⋅(½)4=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Correct
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  6  ⟯
2
(¾)4⋅(¼)2=
6!
(6–2)! ⋅ 2!
(¾)4⋅(¼)2=
6⋅5
2
×
34 ⋅ 12
44 ⋅ 42
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect
⟮  4  ⟯
4
(½)0⋅(½)4=
4!
(4–4)! ⋅ 4!
(½)0⋅(½)4=
1
1
×
1
24
=
1
16
=0.0625 = 6.2%
Incorrect MC

723f_57ae

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)3=
6!
(6–3)! ⋅ 3!
(½)3⋅(½)3=
6⋅5⋅4
3⋅2
×
1
26
=
20
64
=0.3125 = 31.2%
Incorrect
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  9  ⟯
3
(½)6⋅(½)3=
9!
(9–3)! ⋅ 3!
(½)6⋅(½)3=
9⋅8⋅7
3⋅2
×
1
29
=
84
512
=0.1641 = 16.4%
Correct
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  9  ⟯
3
(¾)6⋅(¼)3=
9!
(9–3)! ⋅ 3!
(¾)6⋅(¼)3=
9⋅8⋅7
3⋅2
×
36 ⋅ 13
46 ⋅ 43
=
84 × 729
262144
=0.2336 = 23.4%
Incorrect MC

89b2_e72a

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct
⟮  5  ⟯
3
(¼)2⋅(¾)3=
5!
(5–3)! ⋅ 3!
(¼)2⋅(¾)3=
5⋅4
2
×
12 ⋅ 33
42 ⋅ 43
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  3  ⟯
2
(¾)1⋅(¼)2=
3!
(3–2)! ⋅ 2!
(¾)1⋅(¼)2=
3
1
×
31 ⋅ 12
41 ⋅ 42
=
3 × 3
64
=0.1406 = 14.1%
Incorrect MC

73e4_a592

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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
(¼)6⋅(¾)4=
10!
(10–4)! ⋅ 4!
(¼)6⋅(¾)4=
10⋅9⋅8⋅7
4⋅3⋅2
×
16 ⋅ 34
46 ⋅ 44
=
210 × 81
1048576
=0.0162 = 1.6%
Incorrect
⟮  6  ⟯
4
(¼)2⋅(¾)4=
6!
(6–4)! ⋅ 4!
(¼)2⋅(¾)4=
6⋅5
2
×
12 ⋅ 34
42 ⋅ 44
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  6  ⟯
4
(¾)2⋅(¼)4=
6!
(6–4)! ⋅ 4!
(¾)2⋅(¼)4=
6⋅5
2
×
32 ⋅ 14
42 ⋅ 44
=
15 × 9
4096
=0.0330 = 3.3%
Incorrect
⟮  10  ⟯
6
(¼)4⋅(¾)6=
10!
(10–6)! ⋅ 6!
(¼)4⋅(¾)6=
10⋅9⋅8⋅7
4⋅3⋅2
×
14 ⋅ 36
44 ⋅ 46
=
210 × 729
1048576
=0.1460 = 14.6%
Incorrect
⟮  10  ⟯
6
(½)4⋅(½)6=
10!
(10–6)! ⋅ 6!
(½)4⋅(½)6=
10⋅9⋅8⋅7
4⋅3⋅2
×
1
210
=
210
1024
=0.2051 = 20.5%
Correct MC

6250_d4fd

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  6  ⟯
2
(¾)4⋅(¼)2=
6!
(6–2)! ⋅ 2!
(¾)4⋅(¼)2=
6⋅5
2
×
34 ⋅ 12
44 ⋅ 42
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  8  ⟯
2
(½)6⋅(½)2=
8!
(8–2)! ⋅ 2!
(½)6⋅(½)2=
8⋅7
2
×
1
28
=
28
256
=0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  8  ⟯
2
(¾)6⋅(¼)2=
8!
(8–2)! ⋅ 2!
(¾)6⋅(¼)2=
8⋅7
2
×
36 ⋅ 12
46 ⋅ 42
=
28 × 729
65536
=0.3115 = 31.1%
Incorrect MC

2785_8e65

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
8
(¼)0⋅(¾)8=
8!
(8–8)! ⋅ 8!
(¼)0⋅(¾)8=
1
1
×
10 ⋅ 38
40 ⋅ 48
=
1 × 6561
65536
=0.1001 = 10.0%
Incorrect
⟮  8  ⟯
6
(½)2⋅(½)6=
8!
(8–6)! ⋅ 6!
(½)2⋅(½)6=
8⋅7
2
×
1
28
=
28
256
=0.1094 = 10.9%
Correct
⟮  6  ⟯
2
(¼)4⋅(¾)2=
6!
(6–2)! ⋅ 2!
(¼)4⋅(¾)2=
6⋅5
2
×
14 ⋅ 32
44 ⋅ 42
=
15 × 9
4096
=0.0330 = 3.3%
Incorrect
⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect MC

6f1a_042f

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)2⋅(½)3=
5!
(5–3)! ⋅ 3!
(½)2⋅(½)3=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  8  ⟯
5
(¼)3⋅(¾)5=
8!
(8–5)! ⋅ 5!
(¼)3⋅(¾)5=
8⋅7⋅6
3⋅2
×
13 ⋅ 35
43 ⋅ 45
=
56 × 243
65536
=0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  8  ⟯
3
(½)5⋅(½)3=
8!
(8–3)! ⋅ 3!
(½)5⋅(½)3=
8⋅7⋅6
3⋅2
×
1
28
=
56
256
=0.2188 = 21.9%
Correct
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect MC

33be_cb67

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  5  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect
⟮  7  ⟯
7
(¼)0⋅(¾)7=
7!
(7–7)! ⋅ 7!
(¼)0⋅(¾)7=
1
1
×
10 ⋅ 37
40 ⋅ 47
=
1 × 2187
16384
=0.1335 = 13.3%
Incorrect
⟮  7  ⟯
5
(¼)2⋅(¾)5=
7!
(7–5)! ⋅ 5!
(¼)2⋅(¾)5=
7⋅6
2
×
12 ⋅ 35
42 ⋅ 45
=
21 × 243
16384
=0.3115 = 31.1%
Incorrect
⟮  7  ⟯
2
(½)5⋅(½)2=
7!
(7–2)! ⋅ 2!
(½)5⋅(½)2=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct MC

1d96_7b12

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5=
10!
(10–5)! ⋅ 5!
(½)5⋅(½)5=
10⋅9⋅8⋅7⋅6
5⋅4⋅3⋅2
×
1
210
=
252
1024
=0.2461 = 24.6%
Correct
⟮  10  ⟯
5
(¾)5⋅(¼)5=
10!
(10–5)! ⋅ 5!
(¾)5⋅(¼)5=
10⋅9⋅8⋅7⋅6
5⋅4⋅3⋅2
×
35 ⋅ 15
45 ⋅ 45
=
252 × 243
1048576
=0.0584 = 5.8%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect MC

5a4c_2e72

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  6  ⟯
3
(½)3⋅(½)3=
6!
(6–3)! ⋅ 3!
(½)3⋅(½)3=
6⋅5⋅4
3⋅2
×
1
26
=
20
64
=0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  6  ⟯
3
(¼)3⋅(¾)3=
6!
(6–3)! ⋅ 3!
(¼)3⋅(¾)3=
6⋅5⋅4
3⋅2
×
13 ⋅ 33
43 ⋅ 43
=
20 × 27
4096
=0.1318 = 13.2%
Incorrect MC

7ad5_fe43

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  4  ⟯
2
(¼)2⋅(¾)2=
4!
(4–2)! ⋅ 2!
(¼)2⋅(¾)2=
4⋅3
2
×
12 ⋅ 32
42 ⋅ 42
=
6 × 9
256
=0.2109 = 21.1%
Incorrect
⟮  6  ⟯
4
(½)2⋅(½)4=
6!
(6–4)! ⋅ 4!
(½)2⋅(½)4=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Correct
⟮  6  ⟯
2
(¼)4⋅(¾)2=
6!
(6–2)! ⋅ 2!
(¼)4⋅(¾)2=
6⋅5
2
×
14 ⋅ 32
44 ⋅ 42
=
15 × 9
4096
=0.0330 = 3.3%
Incorrect
⟮  6  ⟯
4
(¼)2⋅(¾)4=
6!
(6–4)! ⋅ 4!
(¼)2⋅(¾)4=
6⋅5
2
×
12 ⋅ 34
42 ⋅ 44
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect MC

89b2_bcfe

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  3  ⟯
2
(¾)1⋅(¼)2=
3!
(3–2)! ⋅ 2!
(¾)1⋅(¼)2=
3
1
×
31 ⋅ 12
41 ⋅ 42
=
3 × 3
64
=0.1406 = 14.1%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect MC

ddec_f933

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
3
(¼)4⋅(¾)3=
7!
(7–3)! ⋅ 3!
(¼)4⋅(¾)3=
7⋅6⋅5
3⋅2
×
14 ⋅ 33
44 ⋅ 43
=
35 × 27
16384
=0.0577 = 5.8%
Incorrect
⟮  4  ⟯
3
(¾)1⋅(¼)3=
4!
(4–3)! ⋅ 3!
(¾)1⋅(¼)3=
4
1
×
31 ⋅ 13
41 ⋅ 43
=
4 × 3
256
=0.0469 = 4.7%
Incorrect
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  7  ⟯
3
(½)4⋅(½)3=
7!
(7–3)! ⋅ 3!
(½)4⋅(½)3=
7⋅6⋅5
3⋅2
×
1
27
=
35
128
=0.2734 = 27.3%
Correct
⟮  7  ⟯
3
(¾)4⋅(¼)3=
7!
(7–3)! ⋅ 3!
(¾)4⋅(¼)3=
7⋅6⋅5
3⋅2
×
34 ⋅ 13
44 ⋅ 43
=
35 × 81
16384
=0.1730 = 17.3%
Incorrect MC

89b2_829f

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Correct
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect
⟮  5  ⟯
2
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect MC

6250_c8c7

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  8  ⟯
6
(¼)2⋅(¾)6=
8!
(8–6)! ⋅ 6!
(¼)2⋅(¾)6=
8⋅7
2
×
12 ⋅ 36
42 ⋅ 46
=
28 × 729
65536
=0.3115 = 31.1%
Incorrect
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  8  ⟯
2
(½)6⋅(½)2=
8!
(8–2)! ⋅ 2!
(½)6⋅(½)2=
8⋅7
2
×
1
28
=
28
256
=0.1094 = 10.9%
Correct
⟮  6  ⟯
6
(½)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  2  ⟯
2
(½)0⋅(½)2=
2!
(2–2)! ⋅ 2!
(½)0⋅(½)2=
1
1
×
1
22
=
1
4
=0.2500 = 25.0%
Incorrect MC

e47f_b592

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
4
(½)5⋅(½)4=
9!
(9–4)! ⋅ 4!
(½)5⋅(½)4=
9⋅8⋅7⋅6
4⋅3⋅2
×
1
29
=
126
512
=0.2461 = 24.6%
Correct
⟮  9  ⟯
5
(¼)4⋅(¾)5=
9!
(9–5)! ⋅ 5!
(¼)4⋅(¾)5=
9⋅8⋅7⋅6
4⋅3⋅2
×
14 ⋅ 35
44 ⋅ 45
=
126 × 243
262144
=0.1168 = 11.7%
Incorrect
⟮  9  ⟯
9
(¼)0⋅(¾)9=
9!
(9–9)! ⋅ 9!
(¼)0⋅(¾)9=
1
1
×
10 ⋅ 39
40 ⋅ 49
=
1 × 19683
262144
=0.0751 = 7.5%
Incorrect
⟮  5  ⟯
4
(¼)1⋅(¾)4=
5!
(5–4)! ⋅ 4!
(¼)1⋅(¾)4=
5
1
×
11 ⋅ 34
41 ⋅ 44
=
5 × 81
1024
=0.3955 = 39.6%
Incorrect
⟮  5  ⟯
4
(¾)1⋅(¼)4=
5!
(5–4)! ⋅ 4!
(¾)1⋅(¼)4=
5
1
×
31 ⋅ 14
41 ⋅ 44
=
5 × 3
1024
=0.0146 = 1.5%
Incorrect MC

ddec_aeb5

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
3
(½)4⋅(½)3=
7!
(7–3)! ⋅ 3!
(½)4⋅(½)3=
7⋅6⋅5
3⋅2
×
1
27
=
35
128
=0.2734 = 27.3%
Correct
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  4  ⟯
4
(½)0⋅(½)4=
4!
(4–4)! ⋅ 4!
(½)0⋅(½)4=
1
1
×
1
24
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  7  ⟯
3
(¾)4⋅(¼)3=
7!
(7–3)! ⋅ 3!
(¾)4⋅(¼)3=
7⋅6⋅5
3⋅2
×
34 ⋅ 13
44 ⋅ 43
=
35 × 81
16384
=0.1730 = 17.3%
Incorrect MC

c4b9_8df9

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(¾)4⋅(¼)2=
6!
(6–2)! ⋅ 2!
(¾)4⋅(¼)2=
6⋅5
2
×
34 ⋅ 12
44 ⋅ 42
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  6  ⟯
2
(½)4⋅(½)2=
6!
(6–2)! ⋅ 2!
(½)4⋅(½)2=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Correct
⟮  6  ⟯
4
(¾)2⋅(¼)4=
6!
(6–4)! ⋅ 4!
(¾)2⋅(¼)4=
6⋅5
2
×
32 ⋅ 14
42 ⋅ 44
=
15 × 9
4096
=0.0330 = 3.3%
Incorrect
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect
⟮  4  ⟯
2
(¾)2⋅(¼)2=
4!
(4–2)! ⋅ 2!
(¾)2⋅(¼)2=
4⋅3
2
×
32 ⋅ 12
42 ⋅ 42
=
6 × 9
256
=0.2109 = 21.1%
Incorrect MC

33be_d249

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(¾)5⋅(¼)2=
7!
(7–2)! ⋅ 2!
(¾)5⋅(¼)2=
7⋅6
2
×
35 ⋅ 12
45 ⋅ 42
=
21 × 243
16384
=0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  5  ⟯
2
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  7  ⟯
2
(½)5⋅(½)2=
7!
(7–2)! ⋅ 2!
(½)5⋅(½)2=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect MC

0700_6508

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)5⋅(¾)3=
8!
(8–3)! ⋅ 3!
(¼)5⋅(¾)3=
8⋅7⋅6
3⋅2
×
15 ⋅ 33
45 ⋅ 43
=
56 × 27
65536
=0.0231 = 2.3%
Incorrect
⟮  8  ⟯
3
(¾)5⋅(¼)3=
8!
(8–3)! ⋅ 3!
(¾)5⋅(¼)3=
8⋅7⋅6
3⋅2
×
35 ⋅ 13
45 ⋅ 43
=
56 × 243
65536
=0.2076 = 20.8%
Incorrect
⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  5  ⟯
3
(½)2⋅(½)3=
5!
(5–3)! ⋅ 3!
(½)2⋅(½)3=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  8  ⟯
5
(½)3⋅(½)5=
8!
(8–5)! ⋅ 5!
(½)3⋅(½)5=
8⋅7⋅6
3⋅2
×
1
28
=
56
256
=0.2188 = 21.9%
Correct MC

5a4c_62e0

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ⋅ 13
43 ⋅ 43
=
20 × 27
4096
=0.1318 = 13.2%
Incorrect
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3=
6!
(6–3)! ⋅ 3!
(½)3⋅(½)3=
6⋅5⋅4
3⋅2
×
1
26
=
20
64
=0.3125 = 31.2%
Correct MC

5a4c_1d36

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ⋅ 13
43 ⋅ 43
=
20 × 27
4096
=0.1318 = 13.2%
Incorrect
⟮  6  ⟯
6
(¼)0⋅(¾)6=
6!
(6–6)! ⋅ 6!
(¼)0⋅(¾)6=
1
1
×
10 ⋅ 36
40 ⋅ 46
=
1 × 729
4096
=0.1780 = 17.8%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  6  ⟯
3
(½)3⋅(½)3=
6!
(6–3)! ⋅ 3!
(½)3⋅(½)3=
6⋅5⋅4
3⋅2
×
1
26
=
20
64
=0.3125 = 31.2%
Correct MC

0700_d0c7

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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 ♀?

⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  8  ⟯
5
(½)3⋅(½)5=
8!
(8–5)! ⋅ 5!
(½)3⋅(½)5=
8⋅7⋅6
3⋅2
×
1
28
=
56
256
=0.2188 = 21.9%
Correct
⟮  8  ⟯
5
(¾)3⋅(¼)5=
8!
(8–5)! ⋅ 5!
(¾)3⋅(¼)5=
8⋅7⋅6
3⋅2
×
33 ⋅ 15
43 ⋅ 45
=
56 × 27
65536
=0.0231 = 2.3%
Incorrect
⟮  8  ⟯
5
(¼)3⋅(¾)5=
8!
(8–5)! ⋅ 5!
(¼)3⋅(¾)5=
8⋅7⋅6
3⋅2
×
13 ⋅ 35
43 ⋅ 45
=
56 × 243
65536
=0.2076 = 20.8%
Incorrect
⟮  5  ⟯
3
(¼)2⋅(¾)3=
5!
(5–3)! ⋅ 3!
(¼)2⋅(¾)3=
5⋅4
2
×
12 ⋅ 33
42 ⋅ 43
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect MC

d05a_c6b1

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)4⋅(½)3=
7!
(7–3)! ⋅ 3!
(½)4⋅(½)3=
7⋅6⋅5
3⋅2
×
1
27
=
35
128
=0.2734 = 27.3%
Incorrect
⟮  10  ⟯
7
(¼)3⋅(¾)7=
10!
(10–7)! ⋅ 7!
(¼)3⋅(¾)7=
10⋅9⋅8
3⋅2
×
13 ⋅ 37
43 ⋅ 47
=
120 × 2187
1048576
=0.2503 = 25.0%
Incorrect
⟮  10  ⟯
7
(½)3⋅(½)7=
10!
(10–7)! ⋅ 7!
(½)3⋅(½)7=
10⋅9⋅8
3⋅2
×
1
210
=
120
1024
=0.1172 = 11.7%
Correct
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect MC

33be_9dde

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
2
(½)5⋅(½)2=
7!
(7–2)! ⋅ 2!
(½)5⋅(½)2=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  5  ⟯
2
(¾)3⋅(¼)2=
5!
(5–2)! ⋅ 2!
(¾)3⋅(¼)2=
5⋅4
2
×
33 ⋅ 12
43 ⋅ 42
=
10 × 27
1024
=0.2637 = 26.4%
Incorrect
⟮  7  ⟯
5
(¼)2⋅(¾)5=
7!
(7–5)! ⋅ 5!
(¼)2⋅(¾)5=
7⋅6
2
×
12 ⋅ 35
42 ⋅ 45
=
21 × 243
16384
=0.3115 = 31.1%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  7  ⟯
2
(¼)5⋅(¾)2=
7!
(7–2)! ⋅ 2!
(¼)5⋅(¾)2=
7⋅6
2
×
15 ⋅ 32
45 ⋅ 42
=
21 × 9
16384
=0.0115 = 1.2%
Incorrect MC

a8ed_67d2

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)0⋅(½)6=
6!
(6–6)! ⋅ 6!
(½)0⋅(½)6=
1
1
×
1
26
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  10  ⟯
4
(½)6⋅(½)4=
10!
(10–4)! ⋅ 4!
(½)6⋅(½)4=
10⋅9⋅8⋅7
4⋅3⋅2
×
1
210
=
210
1024
=0.2051 = 20.5%
Correct
⟮  10  ⟯
4
(¾)6⋅(¼)4=
10!
(10–4)! ⋅ 4!
(¾)6⋅(¼)4=
10⋅9⋅8⋅7
4⋅3⋅2
×
36 ⋅ 14
46 ⋅ 44
=
210 × 729
1048576
=0.1460 = 14.6%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect
⟮  4  ⟯
4
(¼)0⋅(¾)4=
4!
(4–4)! ⋅ 4!
(¼)0⋅(¾)4=
1
1
×
10 ⋅ 34
40 ⋅ 44
=
1 × 81
256
=0.3164 = 31.6%
Incorrect MC

0700_384c

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  3  ⟯
3
(½)0⋅(½)3=
3!
(3–3)! ⋅ 3!
(½)0⋅(½)3=
1
1
×
1
23
=
1
8
=0.1250 = 12.5%
Incorrect
⟮  8  ⟯
5
(½)3⋅(½)5=
8!
(8–5)! ⋅ 5!
(½)3⋅(½)5=
8⋅7⋅6
3⋅2
×
1
28
=
56
256
=0.2188 = 21.9%
Correct
⟮  3  ⟯
3
(¾)0⋅(¼)3=
3!
(3–3)! ⋅ 3!
(¾)0⋅(¼)3=
1
1
×
30 ⋅ 13
40 ⋅ 43
=
1
64
=0.0156 = 1.6%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect MC

1d96_64d0

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  5  ⟯
5
(½)0⋅(½)5=
5!
(5–5)! ⋅ 5!
(½)0⋅(½)5=
1
1
×
1
25
=
1
32
=0.0312 = 3.1%
Incorrect
⟮  5  ⟯
5
(¼)0⋅(¾)5=
5!
(5–5)! ⋅ 5!
(¼)0⋅(¾)5=
1
1
×
10 ⋅ 35
40 ⋅ 45
=
1 × 243
1024
=0.2373 = 23.7%
Incorrect
⟮  10  ⟯
5
(½)5⋅(½)5=
10!
(10–5)! ⋅ 5!
(½)5⋅(½)5=
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
×
15 ⋅ 35
45 ⋅ 45
=
252 × 243
1048576
=0.0584 = 5.8%
Incorrect
⟮  10  ⟯
10
(¼)0⋅(¾)10=
10!
(10–10)! ⋅ 10!
(¼)0⋅(¾)10=
1
1
×
10 ⋅ 310
40 ⋅ 410
=
1 × 59049
1048576
=0.0563 = 5.6%
Incorrect MC

b709_e3a6

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

In this scenario, assume that each child is born independently with the same chance of being either sex. 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  ⟯
5
(¾)2⋅(¼)5=
7!
(7–5)! ⋅ 5!
(¾)2⋅(¼)5=
7⋅6
2
×
32 ⋅ 15
42 ⋅ 45
=
21 × 9
16384
=0.0115 = 1.2%
Incorrect
⟮  5  ⟯
2
(½)3⋅(½)2=
5!
(5–2)! ⋅ 2!
(½)3⋅(½)2=
5⋅4
2
×
1
25
=
10
32
=0.3125 = 31.2%
Incorrect
⟮  7  ⟯
5
(½)2⋅(½)5=
7!
(7–5)! ⋅ 5!
(½)2⋅(½)5=
7⋅6
2
×
1
27
=
21
128
=0.1641 = 16.4%
Correct
⟮  2  ⟯
2
(¾)0⋅(¼)2=
2!
(2–2)! ⋅ 2!
(¾)0⋅(¼)2=
1
1
×
30 ⋅ 12
40 ⋅ 42
=
1
16
=0.0625 = 6.2%
Incorrect
⟮  5  ⟯
2
(¼)3⋅(¾)2=
5!
(5–2)! ⋅ 2!
(¼)3⋅(¾)2=
5⋅4
2
×
13 ⋅ 32
43 ⋅ 42
=
10 × 9
1024
=0.0879 = 8.8%
Incorrect MC

73e4_1f38

Model: Binomial →
⟮  n  ⟯
k
⋅pn-k⋅qk

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

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

⟮  6  ⟯
4
(½)2⋅(½)4=
6!
(6–4)! ⋅ 4!
(½)2⋅(½)4=
6⋅5
2
×
1
26
=
15
64
=0.2344 = 23.4%
Incorrect
⟮  10  ⟯
6
(¼)4⋅(¾)6=
10!
(10–6)! ⋅ 6!
(¼)4⋅(¾)6=
10⋅9⋅8⋅7
4⋅3⋅2
×
14 ⋅ 36
44 ⋅ 46
=
210 × 729
1048576
=0.1460 = 14.6%
Incorrect
⟮  6  ⟯
4
(¼)2⋅(¾)4=
6!
(6–4)! ⋅ 4!
(¼)2⋅(¾)4=
6⋅5
2
×
12 ⋅ 34
42 ⋅ 44
=
15 × 81
4096
=0.2966 = 29.7%
Incorrect
⟮  10  ⟯
4
(¼)6⋅(¾)4=
10!
(10–4)! ⋅ 4!
(¼)6⋅(¾)4=
10⋅9⋅8⋅7
4⋅3⋅2
×
16 ⋅ 34
46 ⋅ 44
=
210 × 81
1048576
=0.0162 = 1.6%
Incorrect
⟮  10  ⟯
6
(½)4⋅(½)6=
10!
(10–6)! ⋅ 6!
(½)4⋅(½)6=
10⋅9⋅8⋅7
4⋅3⋅2
×
1
210
=
210
1024
=0.2051 = 20.5%
Correct