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
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
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F989374ce-3c35-4f4d-b4d6-005e01ca096b%2Faf81e3bb-5e16-4328-aad5-2f2e244981a9%2Fakakc8_processed.jpeg&w=3840&q=75)
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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