13. Suppose that the probability that Bob eats a banana at lunch is 14%. Also, suppose that the probability Bob eats a peach at lunch is 23%. Bob never eats both a banana and a peach at lunch. Find the probability Bob will either eat a banana or a peach at lunch. A. 0.37 B. 0.39 C. 0.41 D. 0.43
13. Suppose that the probability that Bob eats a banana at lunch is 14%. Also, suppose that the probability Bob eats a peach at lunch is 23%. Bob never eats both a banana and a peach at lunch. Find the probability Bob will either eat a banana or a peach at lunch. A. 0.37 B. 0.39 C. 0.41 D. 0.43
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![**Problem 13: Probability of Eating Fruit**
Suppose that the probability that Bob eats a banana at lunch is 14%. Also, suppose that the probability that Bob eats a peach at lunch is 23%. Bob never eats both a banana and a peach at lunch. Find the probability that Bob will either eat a banana or a peach at lunch.
Choices:
- A. 0.37
- B. 0.39
- C. 0.41
- D. 0.43
**Explanation:**
To solve this probability problem, we utilize the principle of mutually exclusive events since Bob never eats both fruits together. The total probability is the sum of the probabilities of the individual events:
\[
P(\text{Banana or Peach}) = P(\text{Banana}) + P(\text{Peach}) = 0.14 + 0.23 = 0.37
\]
Therefore, the correct answer is **A. 0.37**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F989374ce-3c35-4f4d-b4d6-005e01ca096b%2F7e0b093c-1c21-435c-85ad-3409f9327462%2Flvyund4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 13: Probability of Eating Fruit**
Suppose that the probability that Bob eats a banana at lunch is 14%. Also, suppose that the probability that Bob eats a peach at lunch is 23%. Bob never eats both a banana and a peach at lunch. Find the probability that Bob will either eat a banana or a peach at lunch.
Choices:
- A. 0.37
- B. 0.39
- C. 0.41
- D. 0.43
**Explanation:**
To solve this probability problem, we utilize the principle of mutually exclusive events since Bob never eats both fruits together. The total probability is the sum of the probabilities of the individual events:
\[
P(\text{Banana or Peach}) = P(\text{Banana}) + P(\text{Peach}) = 0.14 + 0.23 = 0.37
\]
Therefore, the correct answer is **A. 0.37**.
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