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The question is:

A batch of one hundred items is inspected by testing four randomly selected items. If one of the four is defective, the batch is rejected. What is the probability that the batch is accepted if it contains five defectives?

A simple solution would be to just find out the probability that none of the items are defective.

Let A be the event that the batch will be accepted. Then A = A1 ∩ A2 ∩ A3 ∩ A4, where Ai, i = 1, . . . , 4, is the event that the ith item is not defective. Using the multiplication rule, we have

P(A) = P(A1)P(A2 | A1)P(A3 | A1∩A2)P(A4 | A1∩A2∩A3) = $\frac{95}{100}·\frac{94}{99}·\frac{93}{98}·\frac{92}{97}$ = 0.812.

But we can instead find the probability that none of the defective items are selected. Each item is not selected by probability (1 - 5/100). For selecting 4 items, we get:

P(A) = $(1 - \frac{5}{100}).(1 - \frac{5}{100}).(1 - \frac{5}{100}).(1 - \frac{5}{100})$ = 0.814

Why is the second method giving a different answer?

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Using the second method, you did not subtract the item you have already picked. It should be $$P(A) = (1 - \frac{5}{100})(1 - \frac{5}{99})(1 - \frac{5}{98})(1 - \frac{5}{97}) = 0.812$$

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