11. Suppose that the probability that Bob eats an apple at lunch is 8%. Also, suppose that the probability that Bob eats a pear at lunch is 13%. Also, suppose that the probability that Bob eats both an apple and a pear at lunch is 2%. Find the probability that Bob will either eat an apple or a pear or both at lunch. A. 0.17 В. О.19 С. 0.21 D. 0.23

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**Question 11:**
Suppose that the probability that Bob eats an apple at lunch is 8%. Also, suppose that the probability that Bob eats a pear at lunch is 13%. Also, suppose that the probability that Bob eats both an apple and a pear at lunch is 2%. Find the probability that Bob will either eat an apple or a pear or both at lunch.

Options:
- A. 0.17
- B. 0.19
- C. 0.21
- D. 0.23

**Explanation:**
To find the probability that Bob will eat either an apple or a pear or both, we can use the principle of inclusion-exclusion. The formula is:

\[ P(A \cup P) = P(A) + P(P) - P(A \cap P) \]

Where:
- \( P(A) \) = Probability of eating an apple = 0.08
- \( P(P) \) = Probability of eating a pear = 0.13
- \( P(A \cap P) \) = Probability of eating both an apple and a pear = 0.02

Plug the values into the formula:

\[ P(A \cup P) = 0.08 + 0.13 - 0.02 = 0.19 \]

Thus, the probability that Bob will either eat an apple or a pear or both at lunch is 0.19.

Correct answer: B. 0.19
Transcribed Image Text:**Question 11:** Suppose that the probability that Bob eats an apple at lunch is 8%. Also, suppose that the probability that Bob eats a pear at lunch is 13%. Also, suppose that the probability that Bob eats both an apple and a pear at lunch is 2%. Find the probability that Bob will either eat an apple or a pear or both at lunch. Options: - A. 0.17 - B. 0.19 - C. 0.21 - D. 0.23 **Explanation:** To find the probability that Bob will eat either an apple or a pear or both, we can use the principle of inclusion-exclusion. The formula is: \[ P(A \cup P) = P(A) + P(P) - P(A \cap P) \] Where: - \( P(A) \) = Probability of eating an apple = 0.08 - \( P(P) \) = Probability of eating a pear = 0.13 - \( P(A \cap P) \) = Probability of eating both an apple and a pear = 0.02 Plug the values into the formula: \[ P(A \cup P) = 0.08 + 0.13 - 0.02 = 0.19 \] Thus, the probability that Bob will either eat an apple or a pear or both at lunch is 0.19. Correct answer: B. 0.19
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