a family has 5 children, what is the probability that there

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Mon Sep 10, 2018 9:48 pm

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a family has 5 children, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

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Answer: Option E

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Sep 11, 2018 4:25 am

The problem should make clear that P(boy) = P(girl) = 1/2:

A family has 5 children. If each child is equally likely to be a boy or a girl, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

P(exactly n times) = P(one way) * total possible ways.

Let B = boy and G = girl.

P(one way):
One way to get exactly 3 boys and 2 girls is BBBGG.
P(BBBGG) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.

Total possible ways:
Any arrangement of the letters BBBGG will yield exactly 3 boys and 2 girls.
Thus, to account for all the ways to get exactly 3 boys and 2 girls, the result above needs to be multiplied by the number of ways to arrange BBBGG.
Number of ways to arrange BBBGG = 5!/(3!2!) = 10.

Multiplying the results above, we get:
P(exactly 3 boys and 2 girls) = 10 * 1/32 = 10/32 = 5/16.

The correct answer is E.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

A family has 5 children, what is the probability that there

by fskilnik@GMATH » Tue Sep 11, 2018 4:39 am

A family has 5 children. If each child is equally likely to be a boy or a girl, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

Source: www.GMATinsight.com

\[? = P\left( {3B\,\,{\text{and}}\,\,2G\,\,{\text{among}}\,\,5\,\,{\text{children}}} \right)\]

Imagine 3 BÂ´s and 2 GÂ´s in a row... each sequence is associated with the childrenÂ´s sexuality, from (say) the older to the younger child.

Total number: 2^5 = 32 equiprobable sequences (for instance starting with BBBBB and ending with GGGGG.)

Favorable number: C(5, 3) = 10 , because we must choose among the 5 positions in the sequence, 3 of them to "put" the BÂ´s (and the remaining 2 to "put" the GÂ´s).

\[? = \frac{{C\left( {5,3} \right)}}{{32}} = \frac{{10}}{{32}} = \frac{5}{{16}}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

Last edited by fskilnik@GMATH on Wed Sep 26, 2018 2:18 pm, edited 1 time in total.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8136; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Sep 12, 2018 5:46 pm

GMATinsight wrote:a family has 5 children, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

The number of 5 children of any GMAT is 2 x 2 x 2 x 2 x 2 = 2^5 = 32. The number of ways that the 5 children could be 3 boys and 2 girls is 5!/(3! x 2!) = (5 x 4)/2! = 10. Therefore, the probability is 10/32 = 5/16.

Answer: E

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