AIME 2004 II · 第 2 题

Solution

The probability that Terry picks two red candies is $\frac{10 \cdot 9}{20 \cdot 19} = \frac{9}{38}$, and the probability that Mary picks two red candies after Terry chooses two red candies is $\frac{7\cdot8}{18\cdot17} = \frac{28}{153}$. So the probability that they both pick two red candies is $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}$. The same calculation works for the blue candies.

The probability that Terry picks two different candies is $\frac{20\cdot10}{20\cdot19} = \frac{10}{19}$, and the probability that Mary picks two different candies after Terry picks two different candies is $\frac{18\cdot 9}{18\cdot 17} = \frac{9}{17}$. Thus, the probability that they both choose two different candies is $\frac{10}{19}\cdot\frac{9}{17} = \frac{90}{323}$. Then the total probability is

$$2 \cdot \frac{14}{323} + \frac{90}{323} = \frac{118}{323}$$ and so the answer is $118 + 323 = \boxed{441}$.

In the above calculations, we treated the choices as ordered; that is, Terry chose first one candy, then a second, and so on. We could also solve the problem using unordered choices. The probabilities calculated will all be the same, but the calculations will appear somewhat different. For instance, the probability that Mary chooses two red candies after Terry chose two red candies will have the form $\frac{{8\choose 2}}{{18 \choose 2}}$, and the probability that Terry chooses two different candies will have the form $\frac{{10\choose 1}\cdot{10\choose 1}}{{20\choose2}}$. It is not difficult to see that these yield the same results as our calculations above, as we would expect.