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

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\%`

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
`(\# ways\ to\ succeed)/(\# random\ sets\ of \ 4) =(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
`(\# ways\ to\ succeed)/(\# random\ sets\ of \ 4) =(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\%`