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 the first clerk, and ( A_2 ) be the event that it was checked by the second clerk.

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 that A1 checked the form)
  • ( P(\text{error} | A_2) = 0.02 ) (the probability of error given that A2 checked the form)

We want to find ( P(A_1 | \text{error}) ) and ( P(A_2 | \text{error}) ), the probabilities that the form was checked by A1 and A2, respectively, given that an error was found.

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})} ]

The total probability of error can be expressed as:
[ P(\text{error}) = P(\text{error} | A_1) \cdot P(A_1) + P(\text{error} | A_2) \cdot P(A_2) ]

Now, you can substitute the given values into these formulas to calculate ( P(A_1 | \text{error}) ) and ( P(A_2 | \text{error}) ).

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