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 $P(X=0) = 0.1317$.