This problem involves conditional probabilities. Here's how to solve it step-by-step:

Problem Setup:

• $P(A) = 0.5$, $P(B) = 0.3$, $P(C) = 0.2$ are the probabilities that A, B, or C will get the contract.
• If A gets the contract, $P(E|A) = 0.8$.
• If B gets the contract, $P(E|B) = 0.4$.
• If C gets the contract, $P(E|C) = 0.1$.
• It is known that E got the subcontract. We need to find $P(A|E)$, $P(B|E)$, and $P(C|E)$.

Using Bayes’ Theorem:

The formula for conditional probability is: $P(A|E) = \frac{P(E|A)P(A)}{P(E)}$ Similarly, we can calculate $P(B|E)$ and $P(C|E)$.

Step 1: Find $P(E)$

The total probability of $P(E)$ is: $P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)$ Substitute the given values: $P(E) = (0.8)(0.5) + (0.4)(0.3) + (0.1)(0.2)$ $P(E) = 0.4 + 0.12 + 0.02 = 0.54$

Step 2: Find $P(A|E)$, $P(B|E)$, $P(C|E)$

Now use Bayes' theorem for each case:

1. For A: $P(A|E) = \frac{P(E|A)P(A)}{P(E)} = \frac{(0.8)(0.5)}{0.54} = \frac{0.4}{0.54} = 0.7407$

2. For B: $P(B|E) = \frac{P(E|B)P(B)}{P(E)} = \frac{(0.4)(0.3)}{0.54} = \frac{0.12}{0.54} = 0.2222$

3. For C: $P(C|E) = \frac{P(E|C)P(C)}{P(E)} = \frac{(0.1)(0.2)}{0.54} = \frac{0.02}{0.54} = 0.0370$

Final Results:

• $P(A|E) = 0.7407$
• $P(B|E) = 0.2222$
• $P(C|E) = 0.0370$

Let me know if you'd like a deeper explanation or further assistance!

Here are 5 related questions for further exploration:

1. How would the probabilities change if $P(E|C)$ were higher?
2. Can you determine the probability of E not getting the subcontract?
3. What if the probabilities $P(A)$, $P(B)$, and $P(C)$ were equal?
4. How does Bayes’ theorem apply in other real-life contracting scenarios?
5. Can this problem be modeled using decision trees or diagrams?

Tip: Understanding Bayes’ theorem is critical for problems involving conditional