This problem can be solved using Bayes' Theorem, which allows us to find the probability that a particular assistant processed the mail, given that the mail contains an error.

Let’s define the events:

• $A_1$ is the event that Assistant A processed the mail.
• $A_2$ is the event that Assistant B processed the mail.
• $A_3$ is the event that Assistant C processed the mail.
• $E$ is the event that there is an error in the mail.

We are given:

• $P(A_1) = 0.40$, $P(A_2) = 0.35$, and $P(A_3) = 0.25$ (these are the prior probabilities).
• $P(E | A_1) = 0.04$, $P(E | A_2) = 0.06$, and $P(E | A_3) = 0.03$ (these are the conditional probabilities of making an error, given that a specific assistant processed the mail).

We need to calculate the following conditional probabilities:

• $P(A_1 | E)$, the probability that the mail was processed by Assistant A, given that it has an error.
• $P(A_2 | E)$, the probability that the mail was processed by Assistant B, given that it has an error.
• $P(A_3 | E)$, the probability that the mail was processed by Assistant C, given that it has an error.

Using Bayes' Theorem:

$P(A_i | E) = \frac{P(E | A_i) \cdot P(A_i)}{P(E)}$

Where $P(E)$ is the total probability of there being an error, calculated using the law of total probability:

$P(E) = P(E | A_1) \cdot P(A_1) + P(E | A_2) \cdot P(A_2) + P(E | A_3) \cdot P(A_3)$

Now let’s calculate $P(E)$:

$P(E) = (0.04 \cdot 0.40) + (0.06 \cdot 0.35) + (0.03 \cdot 0.25)$ $P(E) = 0.016 + 0.021 + 0.0075 = 0.0445$

Now, apply Bayes' theorem to find the individual probabilities:

1. For Assistant A: $P(A_1 | E) = \frac{P(E | A_1) \cdot P(A_1)}{P(E)} = \frac{0.04 \cdot 0.40}{0.0445} = \frac{0.016}{0.0445} \approx 0.3596$

2. For Assistant B: $P(A_2 | E) = \frac{P(E | A_2) \cdot P(A_2)}{P(E)} = \frac{0.06 \cdot 0.35}{0.0445} = \frac{0.021}{0.0445} \approx 0.4722$

3. For Assistant C: $P(A_3 | E) = \frac{P(E | A_3) \cdot P(A_3)}{P(E)} = \frac{0.03 \cdot 0.25}{0.0445} = \frac{0.0075}{0.0445} \approx 0.1685$

Final Results:

• $P(A_1 | E) \approx 0.3596$
• $P(A_2 | E) \approx 0.4722$
• $P(A_3 | E) \approx 0.1685$

This means that if a mail is found to have an error, the probability that it was processed by:

• Assistant A is approximately 35.96%
• Assistant B is approximately 47.22%
• Assistant C is approximately 16.85%

Would you like more details on the calculations, or do you have any other questions?

Follow-up Questions:

1. What is the total probability of there being an error in the system?
2. How would the probabilities change if Assistant A’s error rate increased?
3. What is the impact of varying the percentage of mail processed by each assistant?
4. Can we apply this approach to other real-life error-detection systems?
5. How would we compute the probability of no error being made in the entire mail-processing system?

Tip: Bayes' Theorem is useful in decision-making processes when new evidence is introduced, especially for diagnosing or updating beliefs based on observations.