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It seems like you're working on a probability problem involving two events \( A_1 \) and \( A_2 \), where you're given the probabilities for these events and are asked to compute intersections and conditional probabilities using Bayes' Theorem. Let me break down the key points:

### Problem Overview:
- You have the prior probabilities:
  - \( P(A_1) = 0.30 \)
  - \( P(A_2) = 0.70 \)
  - \( P(A_1 \cap A_2) = 0 \) (the events are mutually exclusive).
- The conditional probabilities given are:
  - \( P(B | A_1) = 0.20 \)
  - \( P(B | A_2) = 0.30 \).

### Key Questions:
1. **Are events \( A_1 \) and \( A_2 \) mutually exclusive?** 
   - The answer is "Yes," because \( P(A_1 \cap A_2) = 0 \), which indicates that both events cannot happen at the same time.
   
2. **Computing \( P(A_1 \cap B) \) and \( P(A_2 \cap B) \)**:
   - To compute these, you use the definition of conditional probability:
     \[
     P(A_1 \cap B) = P(B | A_1) \times P(A_1)
     \]
     \[
     P(A_2 \cap B) = P(B | A_2) \times P(A_2)
     \]
     Substituting the given values:
     \[
     P(A_1 \cap B) = 0.20 \times 0.30 = 0.06
     \]
     \[
     P(A_2 \cap B) = 0.30 \times 0.70 = 0.21
     \]
     The correct answers should be \( P(A_1 \cap B) = 0.06 \) and \( P(A_2 \cap B) = 0.21 \).

3. **Computing \( P(B) \)**:
   - Since \( A_1 \) and \( A_2 \) are mutually exclusive, \( P(B) \) can be found by adding the individual contributions of \( A_1 \) and \( A_2 \):
     \[
     P(B) = P(A_1 \cap B) + P(A_2 \cap B) = 0.06 + 0.21 = 0.27
     \]
     This matches the correct value shown in the solution.

4. **Applying Bayes' Theorem to compute \( P(A_1 | B) \) and \( P(A_2 | B) \)**:
   - Bayes' Theorem is applied as follows:
     \[
     P(A_1 | B) = \frac{P(A_1 \cap B)}{P(B)} = \frac{0.06}{0.27} \approx 0.2222
     \]
     \[
     P(A_2 | B) = \frac{P(A_2 \cap B)}{P(B)} = \frac{0.21}{0.27} \approx 0.7778
     \]
     These results match the correct values shown.

It seems the issue lies in part **b** where the values entered for \( P(A_1 \cap B) \) and \( P(A_2 \cap B) \) were incorrect.

Would you like further clarification on any of these steps or more details?

### Follow-up Questions:
1. Why does mutual exclusivity lead to \( P(A_1 \cap A_2) = 0 \)?
2. How would the problem change if \( P(A_1 \cap A_2) \neq 0 \)?
3. Can Bayes' Theorem be used in cases where events are not mutually exclusive?
4. How does the formula for conditional probability apply to this problem?
5. What other types of probability problems can Bayes' Theorem help solve?

### Tip:
Always double-check your calculations, especially when dealing with intersections of events, as small errors in multiplication can lead to larger mistakes in subsequent steps!

The prior probabilities for events A_1 and A_2 are P(A_1) = 0.30 and P(A_2) = 0.70. It is also known that P(A_1 ∩ A_2) = 0. Suppose P(B|A_1) = 0.20 and P(B|A_2) = 0.30.

Explore a probability problem involving mutually exclusive events A_1 and A_2 with given probabilities, where you compute intersections and apply Bayes' Theorem. Detailed solution with steps on how to calculate P(A ∩ B), P(B), and conditional probabilities. Suitable for Grades 11-12.

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