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From a group of 3 boys and 3 girls, 4 children are to be

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From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 22 Jan 2010, 23:42
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A
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D
E

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Question Stats:

76% (01:57) correct 24% (01:38) wrong based on 28 sessions
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
[Reveal] Spoiler: OA
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Re: probability [#permalink] New post 23 Jan 2010, 00:58
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5
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Re: probability [#permalink] New post 23 Jan 2010, 01:24
Expert's post
xcusemeplz2009 wrote:
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5


In this case it's easier to calculate directly what is asked. Equal # of boys and girls out of 4 means 2 boys and 2 girls.

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

\frac{3C2*3C2}{6C4}=\frac{3}{5}

OR: \frac{4!}{2!2!}*\frac{3}{6}*\frac{2}{5}*\frac{3}{4}*\frac{2}{3}=\frac{3}{5}.

Here we are counting the probability of BBGG, this combination can occur in different # of ways: BGBG, GGBB, ... Total # of ways would be the # permutations of the letters BBGG, which is \frac{4!}{2!2!}.

Answer: D.
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Re: Probability problem [#permalink] New post 24 Oct 2010, 23:14
Merging similar topics ....

monirjewel wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal number of boys and girls will be selected?


Total ways to pick children = C(6,4) = 15

Ways to pick 2 boys & 2 girls = C(3,2) x C(3,2) = 9

Probability = 9/15 = 3/5
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Re: probability.... [#permalink] New post 30 Nov 2011, 00:34
There is only 1 way of selecting equal number of boys and girls i.e. 2 boys and 2 girls.

Boys- 3C2 Girls- 3C2
Total possible selections: 6C4

Probability= (3C2*3C2)/6C4 = 3/5
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Re: probability....   [#permalink] 30 Nov 2011, 00:34
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