A lot of 50 items is inspected by the following two-stage plan. A first sample o

Question

A lot of 50 items is inspected by the following two-stage plan. A first sample of 5 items is drawn without replacement. If all are good the lot is passed; if two or more are bad the lot is rejected. If the first sample contains just one bad item, a second sample of 10 more items is drawn without replacement (from the remaining 45 items) and the lot is rejected if two or more of these are bad. Otherwise it is accepted. Suppose there are 10 bad items in the lot. What is the probability that the second sample is drawn and contains more than one bad item? Write down an expression for the probability that the lot is accepted.

Explanation / Answer

a)P(second sample is drawn) = 1 bad item of 5 drawn in a sample of 10 from 50 (noting that 4 good items drawn from 40)=
C(10,1)C(40,4)/C(50,5) = 913900/2118760 = 45695/105938 0.431337197228568

Then, there are 45 items left, with 10-1=9 bad items and 40-4=36 good

P(second sample contains more than one bad item) = 1 - P(0 or 1 bad items)

P(0) = C(9,0)C(36,10)/C(45,10) = 254186856/3190187286 = 1283772/16112057 0.0796777220934608

P(1) = C(9,1)C(36,9)/C(45,10) = 4279240/16112057 0.265592406978203

P(more than one bad item) = 1-1283772/16112057-4279240/16112057= 10549045/16112057 0.654729870928336

Thus, P(second sample is drawn and contains more than 1 bad item) =

45695/105938 * 10549045/16112057 = 482038611275/1706879094466 = 52745225/186768694

0.282409347468051

b)P(lot is accepted) = P(0 bad items in first sample) + P(1 bad item in first sample)P(0 or 1 bad items in second sample|first sample had 1 bad item)

P(0 bad items in first sample) = C(10,0)C(40,5)/C(50,5) = 658008/2118760 = 82251/264845 0.310562782004569

P(1 bad item in first sample)P(0 or 1 bad items in second sample|first sample had 1 bad item) =

45695/105938 * 5563012/16112057 = 254201833340/1706879094466 = 1986790/13340621 0.148927849760517

658008/2118760+1986790/13340621 = 214546163/466921735 0.459490631765086

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