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hard possibility question

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by tanviet » Mon Sep 21, 2009 5:07 am
from a group of 3 boys and girls, 4 children are to be randomly selected. what is the probability that equal number of boys and girls will be selected

a,1/10
b,4/9
c,1/2
d,3/5
e,2/3
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Re: hard possibility question

by Brent@GMATPrepNow » Mon Sep 21, 2009 8:07 am
duongthang wrote:from a group of 3 boys and girls, 4 children are to be randomly selected. what is the probability that equal number of boys and girls will be selected

a,1/10
b,4/9
c,1/2
d,3/5
e,2/3
It's probably best to solve this one using counting techniques.
Probability = (# of ways to select equal number of boys and girls)/(total number of ways to select 4 children)

# of ways to select equal number of boys and girls:
Break this task into 2 stages:
Stage 1: Select 2 boys from 3 boys (can be accomplished in 3C2 ways)
Stage 2: Select 2 girls from 3 girls(can be accomplished in 3C2 ways)
So, the total number of ways to accomplish both stages is 3C2 x 3C2 = 3x3 = 9

Total number of ways to select 4 children:
There are 6 children and we want to randomly select 4 of them.
This can be accomplished in 6C4 ways (15 ways)

So, the probability = 9/15 = 3/5
Brent Hanneson - Creator of GMATPrepNow.com
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