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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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3 Answers

up vote 3 down vote accepted

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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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).

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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 $3/4$.

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Shouldn't the boy,boy case be excluded? There has to be at least one girl. –  Robert Mastragostino May 20 '12 at 15:44
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I think your answer must be wrong for the following reasons: If I change the condition of at least one girl born in March to at least one girl then I am looking for the probability that P(both girls|at least one girl). The possibitities are GB, GG, BG and therefore the probability of being two girls is 1/3. –  adamG May 20 '12 at 15:50
    
Yes, I was mistaken. I think the right answer is $1/3$. –  Stefan Smith May 20 '12 at 16:16
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