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

To find the probabilities that the reservation form was checked by A1 and A2, respectively, given that it has an error, we can use Bayes’ theorem.

Let:

( P(A_1) ) be the probability that a form is checked by clerk A1.
( P(A_2) ) be the probability that a form is checked by clerk A2.
( P(E|A_1) ) be the probability of error given that the form is checked by clerk A1.
( P(E|A_2) ) be the probability of error given that the form is checked by clerk A2.

Given:

We want to find:

( P(A_1|E) ), the probability that the form was checked by A1 given that it has an error.
( P(A_2|E) ), the probability that the form was checked by A2 given that it has an error.

Using Bayes’ theorem:

[
P(A_i|E) = \frac{{P(A_i) \times P(E|A_i)}}{{P(E)}}
]

Where:
[
P(E) = P(A_1) \times P(E|A_1) + P(A_2) \times P(E|A_2)
]

Let’s calculate ( P(E) ) first:

[
P(E) = (0.55 \times 0.03) + (0.45 \times 0.02) = 0.0165 + 0.009 = 0.0255
]

Now, we can calculate ( P(A_1|E) ) and ( P(A_2|E) ):

[
P(A_1|E) = \frac{{0.55 \times 0.03}}{{0.0255}} = \frac{{0.0165}}{{0.0255}} \approx 0.647
]

[
P(A_2|E) = \frac{{0.45 \times 0.02}}{{0.0255}} = \frac{{0.009}}{{0.0255}} \approx 0.353
]

Therefore, the probabilities that the form was checked by A1 and A2, respectively, given that it has an error, are approximately 0.647 and 0.353.