| :) Thanks for taking the time. I've realized a few things, and found it helped to get a bit more formal. Jones has 2 kids. Let A be "he has a girl" and B be "he has a boy born on tuesday." First thing I realized is A and B are NOT independent - this is key. P(A) includes the option of Jones having two girls. But if B is true, then the two girls option isn't on the table anymore, which affects P(A). Realizing this helped me start to better understand what kind of problem we're dealing with. Second was realizing that P(A&B) is not at all the same thing as P(A|B) - the probability of A given B - when A and B aren't independent. The problem is asking for P(A|B), and by the rule of conditional probability: P(A|B) = P(A&B)/P(B) P(B) can be solved for without too much fuss: solve 1-P(!B). For each kid you have 2 genders and 7 days of the week, or 2 * 7 = 14 options. 13 of those are not "Boy & Tuesday." So you have P(!B) is (13/14) * (13/14) = 169/196. P(B) = 1 - 169/196 = 27/196. This leaves us trying to figure out P(A&B). I can't think of any other way to do it other than enumerating all options. We can take a shortcut and just look at all 27 possible scenarios where B is true. This seems to be the method of choice ;) As others have shown, we see that 14 of those satisfy A. So P(A&B) = 14/196. Now, we can solve: P(A|B) = P(A&B)/P(B) = (14/196)/(27/196) = 14/27 So I'm now part of the "math checks out" club. Thanks for all the help people! |