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

Let X will be the number of defective television sets is the sample.

X ={0, 1, 2, 3}

The total number of ways to select 3 items from 15:

$N_t = \frac{15!}{3!(15-3)!} \\ = \frac{13 \times 14 \times 15}{2 \times 3} \\ = 455$

For X=0 (no defective items)

The number of ways to select 3 items from (15-5) = 10 :

$N = \frac{10!}{3!(10-3)!} \\ = \frac{8 \times 9 \times 10}{2 \times 3} \\ = 120 \\ P(X=0) = \frac{120}{455}=0.2637$

For (X=1) (one defective and two non-defective)

The number of ways to select one defective item from 5 :

$N_1 = \frac{5!}{1!(5-1)!} = 5$

The number of ways to select two non-defective items from 10:

$N_2 = \frac{10!}{2!(10-2)!}= \frac{9 \times 10}{2}= 45 \\ P(X=1) = \frac{5 \times 45}{455}=0.4945$

For (X=2) (two defective and one non-defective)

The number of ways to select two defective items from 5 :

$N_1 = \frac{5!}{2!(5-2)!} = \frac{20}{2} = 10$

The number of ways to select one non-defective item from 10:

$N_2 = \frac{10!}{1!(10-1)!}= 10 \\ P(X=2) = \frac{10 \times 10}{455}= 0.2197$

For X=3 (three defective items)

The number of ways to select three defective items from 5 :

$N = \frac{5!}{3!(5-3)!} = \frac{20}{2} = 10 \\ P(X=3) = \frac{10}{455}=0.0219$

The probability distribution of number of defective television sets in the sample