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Natalija [7]
3 years ago
13

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a​ two-st

age 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?
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

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\%

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<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

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