Question

 A large store places its 15 television sets in a clearance sale unknown to anyone, 5 of the television sets are defective. If a customer tests 3 different television sets selected at random, what is the probability distribution of number of defective television sets in the sample?

Answer 1

The probability distribution of the number of defective samples in the sample of 3 taken here is a hypergeometric distribution with N = 15 as the population size, K = 5 as the total number of defectives in population, n = 3 as the sample size.

The probability distribution here is obtained as:

$P(X=x) = \frac{\binom{5}{x} \binom{10}{3-x}}{\binom{15}{3}}, \;x = 0,1,2,3 \\ P(X=0) = \frac{\binom{5}{0} \binom{10}{3}}{\binom{15}{3}} = 0.26373626 \\ P(X=1) = \frac{\binom{5}{1} \binom{10}{2}}{\binom{15}{3}} = 0.49450549 \\ P(X=2) = \frac{\binom{5}{2} \binom{10}{1}}{\binom{15}{3}} = 0.21978022 \\ P(X=3) = \frac{\binom{5}{3} \binom{10}{0}}{\binom{15}{3}} = 0.02197802$

The probability distribution: