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

Solution 1:

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.

Solution 2:

This is leaving mathematics and going into puzzle territory. How come that this statement was made? If you asked the parents "How many girls do you have?" and then asked "can you tell me a month where one of the girls was born"? then the fact that the girl was born in March is irrelevant. On the other hand: I ask a random parent with two children "Do you have a daughter born in March? " and the answer is yes, then it is quite relevant.

Probability of two girls = 1/4. Two girls, at least one born in March (ignoring different lengths of months for simplification), (1/4) * (23 / 144) = 23 / 576.

Probability of one girl = 2/4. One girl born in March = (2/4) * (1/12) = 24 / 576.

Probability that the answer is yes = 47 / 576. Probability of two girls if answer is yes = 23 / 47. Almost 1/2.

So there are many different answers - to find the right one, the exact details of what happened is essential. Here's an alternative set of questions giving the same answers: What was the first month in the year when a child of yours was born? (A: March). Did you have a girl born in March (A: YES).