A car rental agency has 18 compact cars and 12 intermediate size cars. If 4 of the cars are randomly selected for a safety check, what is the probability of getting two of each kind?

Answer 1

I'll show you two ways.

The combinatoric:

choose 2 of 12 and 2 of 18 from 4 of 30

C(12,2)C(18,2)/C(30,4) = 66 * 153/27405 = 10098/27405 = 374/1015 = .3685

The hypergeometric: probability of getting 2 successes of 12 in a sample when 4 successes out of 30 in the population = hypgeom.dist(2,12,4,30,0) = .3685

Using the hypergeometric is especially useful when we want the cumulative sum; then, the final 0 is replaced by a 1 rather than needing multiple combinations in the numerator.

Note that you get the same result for the hypergeometric distribution if you view getting a compact car as a success and you calculate =HYPGEOM.DIST(2,12,4,30,0)