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Having trouble with probability questions

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by sanaa.rizwan » Mon Apr 08, 2013 4:54 pm
NP: QR PS 160

A couple decided to have 4 children, if they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys


A. 3/8
B. 1/4
C.3/16
D. 1/8
E. 1/16
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by Anju@Gurome » Mon Apr 08, 2013 6:43 pm
sanaa.rizwan wrote:A couple decided to have 4 children, if they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys
Total number of outcomes = 2^4 = 16
Total number of favorable outcomes = Number of ways to select either 2 boys OR 2 girls out of 4 = 4C2 = 6
Therefore, required probability = 6/16 = 3/8

The correct answer is A.
Anju Agarwal
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by Deepthi Subbu » Thu Apr 11, 2013 12:31 am
Hi Anju@Gurome,
It is a little confusing on how you arrived at 4C2. Can you please elaborate. Also for these kind of questions, can you state a general rule ?

--Thanks.
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by Anju@Gurome » Thu Apr 11, 2013 1:02 am
Deepthi Subbu wrote:It is a little confusing on how you arrived at 4C2. Can you please elaborate.
We have 4 places to choose for 2 boys and 2 girls.
Now, as we are looking for exactly 2 boys and 2 girls, if we fix 2 places for 2 boys, the remaining 2 places will be occupied by the 2 girls only and vice versa. In other words, once we know the places of two boys, we know the places of two girls.

Hence, number of favorable cases = number of ways to select 2 places out of 4 = 4C2 = 6

Hope that helps.
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Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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