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:
[tex]P(x)=\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}[/tex]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:
[tex]\begin{gathered} n=15 \\ x=0,1 \\ p=3\text{ \%}=\frac{3}{100}=0.03 \end{gathered}[/tex]Since It wishes to guarantee no more than 1 defective item per box, you can set up that:
[tex]P(x\leq1)=\frac{15!}{(15-0)!0!}(0.03)^0(1-0.03)^{15-0}+\frac{15!}{(15-1)!1!}(0.03)^1(1-0.03)^{15-1}[/tex][tex]undefined[/tex]