Guys, the probability that the first athlete wins a medal is 3/12, so there's no way the probability that both win a medal can be higher than 3/12. That is, 3/11 can't be the answer, unless the USA athletes are a whole lot better than the other athletes.

If we are to assume that everyone has an equal probability of winning, then athlete A has a 3/12 chance of winning a medal, and assuming A wins, B has a 2/11 chance of winning a medal. So the probability both win should be

(3/12)*(2/11) = 1/22

Or, you can say that there are 12C3 ways of choosing three medal winners. If both Americans must win, there are 2C2 (that is, one) way of choosing the two Americans, and 10C1 ways of choosing the third medal winner. So the probability should be:

(2C2 * 10C1)/(12C3) = 10/220 = 1/22

Notice that this answer is exactly 1/6 of 3/11, the answer given above. When you list the podiums involving the Americans (AB_, A_B, _AB, etc.) you're assuming order matters, and when you counted the total number of events (12C3), you assumed order did not matter. It doesn't matter whether you assume order matters here as long as you're consistent: you need to count the numerator and the denominator in the same way.