## In a railway reservation office, two clerks are engaged in checking reservation forms. On an average, the first clerk (A1) checks 55% of the forms, while the second (A2) checks the remaining. While A1 has an error rate of 0.03, that of A2 is 0.02. A reservation form is selected at random from the total number of forms checked during a day and is discovered to have an error. Find the probabilities that it was checked by A1 and A2, respectively

Let’s denote the events as follows:
A1 = The form was checked by Clerk A1.
A2 = The form was checked by Clerk A2.
E = The form has an error.

We are given the following probabilities:
P(A1) = Probability that the form was checked by Clerk A1 = 0.55
P(A2) = Probability that the form was checked by Clerk A2 = 0.45
P(E|A1) = Probability of error given that A1 checked the form = 0.03
P(E|A2) = Probability of error given that A2 checked the form = 0.02

We are asked to find the probabilities P(A1|E) and P(A2|E), i.e., the probabilities that the form was checked by A1 and A2, respectively, given that it was found to have an error.

We can use Bayes’ theorem to calculate these probabilities:

P(A1|E) = P(E|A1) * P(A1) / P(E)
P(A2|E) = P(E|A2) * P(A2) / P(E)

To find P(E), we can use the law of total probability:
P(E) = P(E|A1) * P(A1) + P(E|A2) * P(A2)

Let’s plug in the given values:

P(E) = (0.03 * 0.55) + (0.02 * 0.45)
= 0.0165 + 0.009
= 0.0255

Now, let’s calculate P(A1|E) and P(A2|E):

P(A1|E) = (0.03 * 0.55) / 0.0255
= 0.0165 / 0.0255
≈ 0.647

P(A2|E) = (0.02 * 0.45) / 0.0255
= 0.009 / 0.0255
≈ 0.353

So, the probability that the form was checked by A1 given that it has an error is approximately 64.7%, and the probability that it was checked by A2 given that it has an error is approximately 35.3%.