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 (A_1) be the event that the form was checked by Clerk A1, and (A_2) be the event that the form was checked by Clerk A2.

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We are given:
[ P(A_1) = 0.55 ] (the probability that a form is checked by A1)
[ P(A_2) = 0.45 ] (the probability that a form is checked by A2)
[ P(\text{Error} | A_1) = 0.03 ] (the probability of error given A1 checked the form)
[ P(\text{Error} | A_2) = 0.02 ] (the probability of error given A2 checked the form)

We want to find (P(A_1 | \text{Error})) and (P(A_2 | \text{Error})).

Using Bayes’ Theorem:
[ P(A_1 | \text{Error}) = \frac{P(\text{Error} | A_1) \cdot P(A_1)}{P(\text{Error})} ]
[ P(A_2 | \text{Error}) = \frac{P(\text{Error} | A_2) \cdot P(A_2)}{P(\text{Error})} ]

Now, let’s find (P(\text{Error})):
[ P(\text{Error}) = P(\text{Error} | A_1) \cdot P(A_1) + P(\text{Error} | A_2) \cdot P(A_2) ]

Substitute the given values into the equations to find the probabilities.