Question

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

a. Find the probability that the defective item is in box i.
b. Find the probability that the defective item is found in box 1, given that it was actually put in box 1.
c. Find the unconditional probability that the defective item is not found in box 1.
d. Find the conditional probability that the defective item is in box 1, given that it was not found in box 1.
e. Find the conditional probability that the defective item is in box 2, given that it was not found in box 1.
f. Comment on the behavior of these probabilities as n −→ m; as n −→ 0.

Answer 1

a. The probability that the defective item is in box i is 1/k for each box.

b. The probability that the defective item is found in box 1, given that it was actually put in box 1, is 1.

c. The unconditional probability that the defective item is not found in box 1 is (k-1)/k.

d. The conditional probability that the defective item is in box 1, given that it was not found in box 1, is 0.

e. The conditional probability that the defective item is in box 2, given that it was not found in box 1, is [(m-n)/m]^n * (1/k).

f. The behavior of these probabilities as n approaches m and as n approaches 0 depends on the specific values of m, n, and k.

Step-by-step explanation:

a. The probability that the defective item is in box i can be found by considering the number of ways the defective item can be placed in box i out of all possible ways it can be placed in any box. Since there are k boxes and one defective item, the probability is 1/k for each box.

b. The probability that the defective item is found in box 1, given that it was actually put in box 1, is 1. Since we know the defective item was put in box 1, the probability of finding it there is certain.

c. The unconditional probability that the defective item is not found in box 1 can be found by considering the probability that it is found in any other box. Since there are (k-1) other boxes, the probability is (k-1)/k.

d. The conditional probability that the defective item is in box 1, given that it was not found in box 1, can be found using Bayes' theorem. The probability is P(Box 1|Not found in Box 1) = [P(Not found in Box 1|Box 1) * P(Box 1)] / P(Not found in Box 1). Since the defective item was not found in Box 1, the probability is 0.

e. The conditional probability that the defective item is in box 2, given that it was not found in box 1, can be found using Bayes' theorem. The probability is P(Box 2|Not found in Box 1) = [P(Not found in Box 1 | Box 2) * P(Box 2)] / P(Not found in Box 1). The probability of not finding the defective item in box 1 and finding it in box 2 is [(m-n)/m]^n * (1/k).

f. As n approaches m, the probabilities will tend to change. With n approaching 0, the probabilities will also change. However, without knowing the specific values of m, n, and k, it is not possible to make definitive statements about the behavior of these probabilities.