Among eighteen computers in some store, six have defects. Five randomly selected computers are bought for the university lab. Compute the probability that all five computers have no defects.

## Solution

Among eighteen computers in some store, six have defects. That is the probability that a computer have defect is $p = 0.3333$.

Let $X$ denote the number of defective computers out of randomly selected 5 computers.

Here $n = 5$ and $p=0.3333$.

The probability distribution of $X$ is Binomial distribution. That is $X\sim B(5,0.3333)$.

The probability mass function of $X$ is

` $$ \begin{aligned} P(X=x) &= \binom{5}{x} (0.3333)^x (1-0.3333)^{5-x},\\ & \qquad \; x=0,1,\cdots, 5 \end{aligned} $$ `

The probability that all five computers have no defects means number of defective computers $X=0$.

The probability that $X$ is exactly $0$ is

` $$ \begin{aligned} P(X= 0) & =\binom{5}{0} (0.3333)^{0} (1-0.3333)^{5-0}\\ & = 0.1317\\ \end{aligned} $$ `

The probability that all five computers have no defects is $P(X=0) = 0.1317$.