Question

A machine that manufactures automobile parts produces defective parts 12 of the time. If 10 parts produced by this machine are randomly selected, what is the probability that more than 1 of the parts are defective?

Answer 1

For the Binomial distribution:

$$P(x=k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$

$$P(x>1)=1-P(x=0)-P(x=1)$$

$$P(x=0)=\frac{10!}{0!(10-0)!}0.12^0(1-0.12)^{10-0}=0.2785$$

$$P(x=1)=\frac{10!}{1!(10-1)!}0.12^1(1-0.12)^{10-1}=0.3798$$

$$P(x>1)=1-0.2785-0.3798=0.3417$$