Question

Suppose you just received a shipment of 14 televisions 3 of the televisions are defective it to televisions are randomly selected compute the probability that both televisions work what is the probability at least one of the televisions does not work

Answer 1

Given is the binomial problem with n = 2 and probability of defective = 3/13

The probability mass function of binomial distribution is given as

$P(X=x)=\binom{n}{x}p^{x}(1-p)^{n-x}$

1) Probability that both televisions work

Probability that both televisions work is given as probability that there is no defective piece.

$P(X=0)=\binom{2}{0}(3/13)^{0}(1-3/13)^{2-0}=1 \times 1 \times (10/13)^2 = 0.5917$

2) Probability at least one of the televisions does not work

Probability that at least one of the televisions does not work is given as

$P(X\ge 1)=1-P(X=0) = 1 - 0.5917=0.4083$