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 nd

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.

To find the conditional probability in both cases, we need to use the concept of conditional probability and Bayes' theorem. The conditional probability is the probability of an event A occurring, given that event B has already occurred.

In this case, event A is finding the defective item in a particular box, and event B is not finding the defective item in box 1.

Let's calculate the conditional probabilities step by step:

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

To do this, we need to use Bayes' theorem, which states:

P(A|B) = (P(B|A) * P(A)) / P(B)

Where:
P(A|B) is the conditional probability of A given B,
P(B|A) is the conditional probability of B given A,
P(A) is the probability of A occurring,
P(B) is the probability of B occurring.

In this scenario:
A: The defective item is in box 1.
B: The defective item is not found in box 1.

Now let's break down the terms:

P(A|B) = Probability that the defective item is in box 1 given that it is not found in box 1.
P(B|A) = Probability that the defective item is not found in box 1, given that it is in box 1.
P(A) = Probability that the defective item is in box 1.
P(B) = Probability that the defective item is not found in box 1.

We can calculate P(A|B) as follows:

P(A|B) = (P(B|A) * P(A)) / P(B)

P(B|A) = Probability that the item is not found in box 1, given that it is in box 1. Since all boxes contain m items and only one item is defective, the probability of not finding the defective item in box 1 is (m-1)/m.

P(A) = Probability that the defective item is in box 1. Since there are k boxes and only one of them contains the defective item, the probability is 1/k.

P(B) = Probability that the item is not found in box 1. This is the sum of the probabilities of not finding the defective item in each box. For each box, the probability of not finding the defective item is (m-1)/m. Since there are k boxes, we multiply this probability by k to get the overall probability.

Therefore,

P(B) = k * (m-1)/m

Putting it all together,

P(A|B) = ((m-1)/m * 1/k) / (k * (m-1)/m)
= 1/k

So, the conditional probability that the defective item is in box 1, given that it is not found in box 1, is 1/k.

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

To calculate this probability, we can modify the previous calculation by updating the values of P(A) and P(B|A):

P(A) = Probability that the defective item is in box 2. Since there are k boxes and only one of them contains the defective item, the probability is 1/k.

P(B|A) = Probability that the item is not found in box 1, given that it is in box 2. Since we are assuming the defective item is in box 2, the probability of not finding it in box 1 is still (m-1)/m.

Using the updated values in the formula:

P(A|B) = (P(B|A) * P(A)) / P(B)

P(A|B) = ((m-1)/m * 1/k) / (k * (m-1)/m)
= 1/k

So, the conditional probability that the defective item is in box 2, given that it is not found in box 1, is also 1/k.

In conclusion, the conditional probability that the defective item is in a specific box, given that it is not found in box 1, is the same for all boxes and is given by 1/k.

(a) "Given that it is not found in box 1" means that it is not in box 1, so probability of finding it in box 1 is zero.

(b) Conditional probability:
P(A|B)=P(A∩B)/P(B).
Event A: defective in box 2
Event B: defective not in box 1
P(A)=1 box out of k = 1/k
P(B)=not in one of k boxes = (k-1)/k
P(A∩B)=P(A)=1/k
because if it is in box 2, then it implies that it is not in box 1,
i.e. event A is contained in event B.

P(A|B)=P(A∩B)/P(B)
=(1/k)÷((k-1)/k)
=1/(k-1)

Actually this result can be obtained by intuition, because out of k boxes, knowing that item is not in box 1, there are only (k-1) boxes left, hence P(A|B)=1/(k-1)!