I have a problem, I'm learning probability at the moment (I'm a programmer) and starting I have this:
(Source: Minka.) My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities—two boys or two girls—have probabilities 1/4 and 1/4.
a. Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl?
b. Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl?
Now my reasoning is:
BB = 1/4 = 0.25 BG = 1/4 = 0.25 GB = 1/4 = 0.25 GG = 1/4 = 0.25 So for a., the probability of G I get it just by summing p(B,G) + p(G,B) = 0.5
And for b. p(G|B) = p(G,B)/p(B) = 0.5/0.5 = 1 that is wrong but I'm not getting why.
Best Answer
For (a), a simple way to look at is that you've reduced your probability space to only the combinations that have at least one boy:
BB = 1/3 BG = 1/3 GB = 1/3 GG is no longer a possibility based on the fact that your neighbor said he had at least one boy. Of the possibilities remaining, you're left with a 2/3 probability that he has a girl. The information he gave you reduced the probability of him having a girl from 3/4 to 2/3. Formally, this can be shown as follows: $$P(At least one girl|At least one boy) = frac{P(At least one girl cap At least one boy)}{P(At least one boy)} $$ From your original box, we can see the probability of having at least one boy and at least one girl is BG + GB = 0.25 + 0.25 = 0.5, but we need to divide by the probability of at least one boy, which is BB + BG + GB = 0.25 + 0.25 + 0.25 = 0.75, so we get $frac{tfrac{1}{2}}{tfrac{3}{4}} = frac{2}{3}$.
For (b), now that we've seen a boy, the only uncertainty remaining is the gender of the other child, and given no other information, the probability of the other child being female is 1/2, which is the answer.
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