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 Post subject: Probablity
 Post Posted: Wed Nov 09, 2011 11:26 pm 
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Course Students


Posts: 9
Can you explain me the following question throught the slot method?

A couple decides to have 4 children. If they succeed in havng 4 children and each child is equally likely to be a boy or a girl, what is the probablity that they will have exactly 2 boys and 2 girls?

A)3/8
B)1/4
C)3/16
D) 1/8
E)1/16


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 Post subject: Re: Probablity
 Post Posted: Thu Nov 10, 2011 8:58 am 
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Students


Posts: 1
Probability of having a girl or a boy is 1/2.

Because the question asks for 2 boys and 2 girls,the number of possible combinations is 4!/(2! * 2!) = 6

Consider 2 girls and 2 boys as same.

For each combination,probability is 1/2*1/2*1/2*1/2 = 1/16

therefore for all the combinations value will be 6*1/16 = 3/8.

Hence the ans is A


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 Post subject: Re: Probablity
 Post Posted: Wed Nov 23, 2011 6:18 am 
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ManhattanGMAT Staff


Posts: 7146
swapna wrote:
A couple decides to have 4 children. If they succeed in havng 4 children and each child is equally likely to be a boy or a girl, what is the probablity that they will have exactly 2 boys and 2 girls?

A)3/8
B)1/4
C)3/16
D) 1/8
E)1/16


the solution posted above this one works. if you don't think of that, you can also just list out all the possibilities for two boys and two girls:
bbgg
bgbg
bggb
gbbg
gbgb
ggbb
that's six different possibilities. the total number of ways in which the two children can be born is 2 x 2 x 2 x 2 (from the slot method) = 16, so the probability that you want is 6/16 = 3/8.


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