You have 35 batteries, 4 of 35 are defective, take a sample of 7.. Probability that exactly 2 are defective?

Answer 1

We want to choose 2 of 4 defective batteries and 5 of 31 good batteries, so the number of ways to do it is
C(4, 2) . C(31, 5) = 6 * 169911 = 1019466.
The total number of possibilities is C(35, 7) = 6724520. So the probability is 1019466 / 6724520 = 0.1516 to 4 d.p.