From 10 men, 5 women, 4-member committees with at least 1 woman are formed.

Question

From 10 men, 5 women, 4-member committees with at least 1 woman are formed. Find the probability of more women than men.

Accepted Answer

$$P(3M,1W)+P(2M,2W)+P(1M,3W)+P(0M,4W) = \frac{600+450+100+5}{1365}=\frac{1155}{1365}$$

$$P(\text{more women}) = \frac{P(1M,3W)+P(0M,4W)}{1155/1365} = \frac{105}{1155}=\frac{1}{11}$$

Solutions · 1