silemus.

51Hypergeometric probability

A box has 10 items, 3 of them defective. 4 are chosen at random without replacement. Find P(exactly 1 is defective)\displaystyle P(\text{exactly 1 is defective}).

Use the hypergeometric setup: choose 1 defective from 3, and 3 good ones from 7, divided by all ways to choose 4 from 10. That gives C(3,1)·C(7,3)/C(10,4) = 105/210 = 1/2.

How good is this answer?