A committee of 4 people is chosen at random from 5 men and 4 women. What is the probability that the committee has exactly 2 men and 2 women?

The 4 members can be chosen in $C(9,4)=126$ ways. Picking 2 of the 5 men and 2 of the 4 women gives $C(5,2)\times C(4,2)=10\times 6=60$ committees, so the probability is 60/126 = 10/21. Adding the two counts instead of multiplying them gives 16/126 = 8/63, and 20/63 counts the committees with exactly 3 men.