Of the parts produced by a particular machine, 0.9% are defective. If a random sample of 7 parts produced by this machine contains 2 or more defective parts, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on this sampling plan.

#### Solution

Let $X$ denote the number of defectives within 7 selected parts.

Let $p=0.09$ be the proportion of defectives. Then

$X\sim B(7,0.09)$.

The probability mass function of $X$ is

` $$ \begin{aligned} P(X=x) &= \binom{7}{x} (0.09)^x (1-0.09)^{7-x},\\ & \qquad \; x=0,1,\cdots, 7. \end{aligned} $$ `

The machine is shut down for repairs if 2 or more parts are defective. That is $X\geq 2$.

Thus the probability that the machine will be shut down for repairs based on this sampling plan is

` $$ \begin{aligned} P(X\geq 2) &= 1- P(X < 2)\\ &= 1- P(X\leq 1)\\ &= 1- \bigg(P(X=0) + P(X=1)\bigg)\\ &=1-\bigg(\binom{7}{0} (0.09)^{0} (1-0.09)^{7-0}+\binom{7}{1} (0.09)^{1} (1-0.09)^{7-1}\bigg)\\ & = 1-0.8745\\ &= 0.1255 \end{aligned} $$ `