Question

The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

There are $\binom{9}{3} = 84$ equally likely sets of three sleepers. Exactly two sleepers come from the same country when one country supplies exactly two of them and the third sleeper comes from a different country: $\binom{4}{2}(2 + 3) = 30$ ways with the pair from the United States, $\binom{3}{2}(2 + 4) = 18$ with the pair from Canada, and $\binom{2}{2}(3 + 4) = 7$ with the pair from Mexico.

The probability is $\frac{30 + 18 + 7}{84} = \frac{55}{84},$ already in lowest terms, so $m + n = 55 + 84 = 139.$

2015 AIME I Problem 2

Solution:

Problem 2 in Other Years