Question

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a​ two-stage one. Boxes of 23 items are readied for​ shipment, and a sample of 4 items is tested for defectives. If any defectives are​ found, the entire box is sent back for​ 100% screening. If no defectives are​ found, the box is shipped. ​(a) What is the probability that a box containing 3 defectives will be​ shipped? ​(b) What is the probability that a box containing only 1 defective will be sent back for​ screening?

Answer 1

Answer:

a) The probability that a box containing 3 defectives will be​ shipped is $51.74\%$

b) The probability that a box containing only 1 defective will be sent back for​ screening is $17.39\%$

Step-by-step explanation:

Hi

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so $\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)$, therefore the probability that a box containing 3 defectives will be shipped is $P(S)=54.71\%$

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so $\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315$

Then we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)$, therefore the probability that a box containing 1 defective will be shipped is $P(S)=82.60\%$

Finally the probability that a box containing only 1 defective will be sent back for​ screening will be $P(BS)=1-P(S)=1-0.8260=0.1739=17.39\%$