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Message: <blockquote data-quote="orsal" data-source="post: 1355999" data-attributes="member: 16016">Well, that's not my principal field, but I've taught now at two universities where statistics is part of the math department, and at both schools I taught introductory stats classes as part of my teaching load.OK. I won't bother with spoiler tags, since I don't think anybody is going to get anything from this explanation by casually glancing over it.There are four possibilities for a two-child family: BB, BG, GB, GG. All of them are equally likely (probability 1/4). If you insist on not distinguishing BG from GB, just call them both "1 of each", that's OK, but you no longer have uniform probability -- now P(1 of each)=1/2, P(2 boys)=1/4, P(2 girls)=1/4.Suppose you know there is at least one boy. Then you eliminate (2 girls) as a possibility. You're looking for conditional probabilities: P(2 boys|at least 1 boy), which is read "probability of two boys, given at least one boy", and P(1 of each| at least one boy). To calculate conditional probabilities, you divide by P(at least one boy), which is 1/4 + 1/2 = 3/4. This ensures that the probabilities still add up to 1. Thus we get P(2 boys | at least 1 boy)=(1/4)/(3/4)=1/3, and P(1 of each | at least one boy)=(1/2)/(3/4)=2/3.This is the correct calculation on the assumption that if there is at least one boy in the family, we will learn that, but will never learn any more. Pielorinho, and I, and several other people, did this interpretation.On the other hand, suppose that we were destined to learn the sex of one of the kids, but it could be either. It could be that we would be told the sex of whichever one was older, or whichever one we happened to meet first, or whichever one our friend randomly decided to tell us about. ThenP(we learn of a boy | 2 boys) =1so P(there are 2 boys, and we learn of at least one boy)=(1/4)(1)=1/4But P(we learn of a boy | one of each) =1/2so P(there is one of each, and we learn of at least one boy)=(1/2)(1/2)=(1/4)and P(we learn of a boy | 2 girls) =0so P(there are two girls, and we learn of a boy)=(1/4)(0)=0These calculations all come from the formula P(A and B)=P(A)P(B|A).Altogether, P(we learn of a boy)=(1/4)+(1/4)=(1/2)so we calculate the conditional probabilities asP(2 boys | we learn of a boy) = (1/4)/(1/2) = (1/2)P(one of each | we learn of a boy) = (1/4)/(1/2) = (1/2)which justifies Merak Spielman's answer.</blockquote>

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