# Modified two child problem. Find the probability that both are girls, given that at least one is a girl born in March.

A family has two children. Assume that birth month is independent of gender, with boys and girls equally likely and all months equally likely, and assume that the elder child’s characteristics are independent of the younger child’s characteristics).

What is the probability that both are girls, given that at least one is a girl who was born in March.

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Looks like we owe this variation to Gary Foshee.

In the simpler (no month information) formulation of the two-child problem we have BG, GB, GG as the possible genders of the children for a 1/3 likelihood that both are girls. Here we can again enumerate the possibilities to find (counterintuitively) that the likelihood that both are girls has increased!

We have B[all months]G[march], G[march],B[all months], G[all months]G[march], and G[march]G[all months] as the possible outcomes - but have double-counted the case of G[march]G[march] - which gives us $\frac{2*12-1}{4*12-1} = \frac{23}{47}$ or just shy of a 50% chance that both children are girls.

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The fact that one of the girls were born in March is irrelevant. There are four equally likely possiblities: older=girl, younger=boy; older=girl,younger=girl; older=boy, younger=girl; and older=boy, younger=boy. In 3 out of four cases, one of the children is a girl. So the probability is $3/4$.
Yes, I was mistaken. I think the right answer is $1/3$. –  Stefan Smith May 20 '12 at 16:16