A committee of 3 is to be chosen from 5 men and 4 women. If the selection is made at random, what is the probability that the committee consists of 2 men and 1 woman?

The number of ways to choose 2 men from 5 is C(5,2) = 10, and to choose 1 woman from 4 is C(4,1) = 4, giving 10 × 4 = 40 favourable outcomes. The total number of ways to choose 3 from 9 is C(9,3) = 84. The probability is 40/84 = 10/21.