Question

A manufacturing process produces, on the average, 3% defective items. The company ships 15 items in each box and wishes to guarantee no more than 1 defective item per box. If this guarantee accompanies each box, what is the probability that the box will fail to satisfy the guarantee?

Answer 1

Given that:

- The manufacturing process produces on average 3% defective items.

- The company ships 15 items in each box.

- It wishes to guarantee no more than 1 defective item per box.

You need to use the following Binomial Distribution Formula in order to find the probability that the box will fail to satisfy the​ guarantee:

`P(x)=(n!)/((n-x)!x!)p^x(1-p)^(n-x)`

Where "n" is the number of trials, "x" is the number of successes desired, and "p" is the probability of getting a success in one trial.

In this case:

`\begin{gathered} n=15 \\ x=0,1 \\ p=3\text{ \%}=(3)/(100)=0.03 \end{gathered}`

Since It wishes to guarantee no more than 1 defective item per box, you can set up that:

`P(x\leq1)=(15!)/((15-0)!0!)(0.03)^0(1-0.03)^(15-0)+(15!)/((15-1)!1!)(0.03)^1(1-0.03)^(15-1)`