Question

After packing k boxes (numbered 1,2,⋯,k ) of m items each, workers discovered that one defective item had slipped in among the k×m items packed. In an attempt to find the defective item, they randomly select n items from each box and examine these.

(a) Find the conditional probability that the defective item is in box 1, given that it is not found in box 1.
(b) Find the conditional probability that the defective item is in box 2 , given that it is not found in box 1.

Answer 1

The conditional probabilities that the defective item is in box 1 or box 2, given that it is not found in box 1, are both equal to 1/(k(m-n)), assuming n items are inspected in each box and there are m items per box.

Step-by-step explanation:

To solve these problems, we need to understand conditional probability and work with the given information.

(a) The conditional probability that the defective item is in box 1, given that it is not found in box 1 after examining n items, is calculated as follows:

P(Defective in Box 1 | Not found in Box 1) = P(Defective in Box 1) / P(Not found in Box 1)

The probability that the defective item is in box 1 is 1/k, and the probability that it is not found in box 1 after examining n out of m items is the binomial coefficient (m-n) choose 1 divided by m choose 1. Hence:

P(Defective in Box 1 | Not found in Box 1) = (1/k) / [(m-n)/m] = 1/(k(m-n))

(b) The conditional probability that the defective item is in box 2, given that it is not found in box 1, is the same as that for box 1 because the problem is symmetric with respect to all boxes,

So, P(Defective in Box 2 | Not found in Box 1) = 1/(k(m-n)) as well.