Answer to Question #96624 in Statistics and Probability for Brandyn Popejoy

Question

Suppose you just received a shipment of fifteen televisions. Four of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

Answer 1

In how many ways can 2 televisions be selected from 15?

$$\binom{15}{2}={15! \over 2!(15-2)!}={15\cdot 14 \over 1\cdot 2}=105$$

In how many ways can 0 defective televisions be selected from 4?

$$\binom{4}{0}={4! \over 0!(4-0)!}=1$$

In how many ways can 2 non-defective televisions be selected from 11?

$$\binom{11}{2}={11! \over 2!(11-2)!}={11\cdot 10 \over 1\cdot 2}=55$$

Compute the probability that both televisions work

$$P(both\ work)={\binom{4}{0}\binom{11}{2} \over \binom{15}{2}}={1\cdot 55 \over 105}={11 \over 21}\approx0.5238$$

What is the probability at least one of the two televisions does not​ work?

$$P(at\ least \ 1\ does\ not\ work)=1-P(both\ work)=$$ $$=1-{11 \over 21}={10 \over 21}\approx0.4762$$