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I'm stuck on the following probability problem: parents keep having children until they have one girl, at which point they stop; and babies are girls with probability 0.49. If we select a child uniformly at random (from the entire population of children) what's the probability he or she has exactly one sibling?

If a couple has only $2$ children it means that they had first a boy and then a girl. So why the probability asked is not $0.51 \cdot 0.49$? What I am missing?

It seems that the correct answer is $2 \cdot 0.51 \cdot (0.49)^2$, but I don't know how to get this result. Thanks for your help.

N. F. Taussig
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Ana
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1 Answers1

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$0.51\times 0.49$ is the probability that a randomly-selected family has exactly two children. This is not the same as the probability that a randomly-selected child has exactly one sibling, because if you select a child at random you are more likely to select from larger families.

Write $m$ for the average number of children in a family. If you have a large number $n$ of families you have about $mn$ children. About $0.51\times 0.49n$ families have two children, so about $2\times0.51\times 0.49n$ children have exactly one sibling. So the probability you're looking for is $2\times0.51\times 0.49/m$. Do you know how to work out the value of $m$?

Especially Lime
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  • Thank you very much for all the answers. I guess that the number of children in a family is a Geometric Distribution the success being "having a girl" with probability p=0.49. So the expected value (m) is 1/p=1/0.49 and we get the result. – Ana May 24 '18 at 11:49
  • @Ana yes, that's right. – Especially Lime May 24 '18 at 12:21