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The diagram you provided seems to depict a probability tree with events \( A \) and \( B \) leading to outcomes \( C \) and \( D \). The probabilities on the branches represent conditional probabilities.

Here’s how to approach solving for the outcomes:

1. **\( P(A \cap C) \)**: This represents the probability of going through event \( A \) and then to event \( C \). To find this:
   \[
   P(A \cap C) = P(A) \times P(C|A)
   \]
   From the diagram:
   \[
   P(A) = 0.3, \quad P(C|A) = 0.4
   \]
   Therefore,
   \[
   P(A \cap C) = 0.3 \times 0.4 = 0.12
   \]

2. **\( P(A \cap D) \)**: This is the probability of going through event \( A \) and then to event \( D \). It follows the same rule:
   \[
   P(A \cap D) = P(A) \times P(D|A)
   \]
   From the diagram:
   \[
   P(A) = 0.3, \quad P(D|A) = 1 - 0.4 = 0.6
   \]
   So,
   \[
   P(A \cap D) = 0.3 \times 0.6 = 0.18
   \]

3. **\( P(B \cap C) \)**: This represents the probability of going through event \( B \) and then to event \( C \):
   \[
   P(B \cap C) = P(B) \times P(C|B)
   \]
   From the diagram:
   \[
   P(B) = 1 - 0.3 = 0.7, \quad P(C|B) = 0.1
   \]
   Therefore,
   \[
   P(B \cap C) = 0.7 \times 0.1 = 0.07
   \]

4. **\( P(B \cap D) \)**: This is the probability of going through event \( B \) and then to event \( D \):
   \[
   P(B \cap D) = P(B) \times P(D|B)
   \]
   From the diagram:
   \[
   P(B) = 0.7, \quad P(D|B) = 1 - 0.1 = 0.9
   \]
   So,
   \[
   P(B \cap D) = 0.7 \times 0.9 = 0.63
   \]

### Final Probabilities:
- \( P(A \cap C) = 0.12 \)
- \( P(A \cap D) = 0.18 \)
- \( P(B \cap C) = 0.07 \)
- \( P(B \cap D) = 0.63 \)

Would you like any further details on these calculations?

### Follow-up Questions:
1. How would you calculate \( P(C) \) and \( P(D) \) from this tree?
2. What would the total probability \( P(C) + P(D) \) sum up to?
3. How could we extend this tree diagram if there were additional outcomes beyond \( C \) and \( D \)?
4. Could this method be applied to real-life decision-making scenarios, such as medical testing or financial investments?
5. How would the probabilities change if event \( A \) or \( B \) were altered?

### Tip:
Always ensure that the sum of probabilities at each branch adds up to 1.

The diagram you provided seems to depict a probability tree with events A and B leading to outcomes C and D. The probabilities on the branches represent conditional probabilities. Here’s how to approach solving for the outcomes:

1. P(A ∩ C): This represents the probability of going through event A and then to event C. To find this: P(A ∩ C) = P(A) × P(C|A). From the diagram: P(A) = 0.3, P(C|A) = 0.4. Therefore, P(A ∩ C) = 0.3 × 0.4 = 0.12.

2. P(A ∩ D): This is the probability of going through event A and then to event D. It follows the same rule: P(A ∩ D) = P(A) × P(D|A). From the diagram: P(A) = 0.3, P(D|A) = 1 - 0.4 = 0.6. So, P(A ∩ D) = 0.3 × 0.6 = 0.18.

3. P(B ∩ C): This represents the probability of going through event B and then to event C: P(B ∩ C) = P(B) × P(C|B). From the diagram: P(B) = 1 - 0.3 = 0.7, P(C|B) = 0.1. Therefore, P(B ∩ C) = 0.7 × 0.1 = 0.07.

4. P(B ∩ D): This is the probability of going through event B and then to event D: P(B ∩ D) = P(B) × P(D|B). From the diagram: P(B) = 0.7, P(D|B) = 1 - 0.1 = 0.9. So, P(B ∩ D) = 0.7 × 0.9 = 0.63.

### Final Probabilities:
- P(A ∩ C) = 0.12
- P(A ∩ D) = 0.18
- P(B ∩ C) = 0.07
- P(B ∩ D) = 0.63

Learn how to calculate conditional probabilities using a probability tree diagram with events A and B leading to outcomes C and D. This problem involves the multiplication rule for independent events and is suitable for students in Grades 10-12.

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