**Question** In a certain group of 63 students, 24 are taking economics, 19 are taking precalculus, and 9 are taking both. What is the probability that a randomly chosen student is taking either economics or precalculus? Give your answer as a fraction. **Solution Explanation** To find the probability that a student is taking either economics or precalculus, use the formula for the union of two sets. The formula is: \[ P(E \cup P) = P(E) + P(P) - P(E \cap P) \] where: - \( P(E) \) is the probability of a student taking economics. - \( P(P) \) is the probability of a student taking precalculus. - \( P(E \cap P) \) is the probability of a student taking both subjects. Plugging in the numbers: - 24 students are taking economics, so \( P(E) = \frac{24}{63} \). - 19 students are taking precalculus, so \( P(P) = \frac{19}{63} \). - 9 students are taking both, so \( P(E \cap P) = \frac{9}{63} \). Now compute: \[ P(E \cup P) = \frac{24}{63} + \frac{19}{63} - \frac{9}{63} = \frac{34}{63} \] Therefore, the probability that a randomly chosen student is taking either economics or precalculus is \( \frac{34}{63} \).
**Question** In a certain group of 63 students, 24 are taking economics, 19 are taking precalculus, and 9 are taking both. What is the probability that a randomly chosen student is taking either economics or precalculus? Give your answer as a fraction. **Solution Explanation** To find the probability that a student is taking either economics or precalculus, use the formula for the union of two sets. The formula is: \[ P(E \cup P) = P(E) + P(P) - P(E \cap P) \] where: - \( P(E) \) is the probability of a student taking economics. - \( P(P) \) is the probability of a student taking precalculus. - \( P(E \cap P) \) is the probability of a student taking both subjects. Plugging in the numbers: - 24 students are taking economics, so \( P(E) = \frac{24}{63} \). - 19 students are taking precalculus, so \( P(P) = \frac{19}{63} \). - 9 students are taking both, so \( P(E \cap P) = \frac{9}{63} \). Now compute: \[ P(E \cup P) = \frac{24}{63} + \frac{19}{63} - \frac{9}{63} = \frac{34}{63} \] Therefore, the probability that a randomly chosen student is taking either economics or precalculus is \( \frac{34}{63} \).
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Question
![**Question**
In a certain group of 63 students, 24 are taking economics, 19 are taking precalculus, and 9 are taking both. What is the probability that a randomly chosen student is taking either economics or precalculus? Give your answer as a fraction.
**Solution Explanation**
To find the probability that a student is taking either economics or precalculus, use the formula for the union of two sets.
The formula is:
\[ P(E \cup P) = P(E) + P(P) - P(E \cap P) \]
where:
- \( P(E) \) is the probability of a student taking economics.
- \( P(P) \) is the probability of a student taking precalculus.
- \( P(E \cap P) \) is the probability of a student taking both subjects.
Plugging in the numbers:
- 24 students are taking economics, so \( P(E) = \frac{24}{63} \).
- 19 students are taking precalculus, so \( P(P) = \frac{19}{63} \).
- 9 students are taking both, so \( P(E \cap P) = \frac{9}{63} \).
Now compute:
\[ P(E \cup P) = \frac{24}{63} + \frac{19}{63} - \frac{9}{63} = \frac{34}{63} \]
Therefore, the probability that a randomly chosen student is taking either economics or precalculus is \( \frac{34}{63} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F140bf091-6917-4391-bf36-7a6766f112e3%2F978f45ae-4a7a-472c-8e10-8bdbf4b9dc80%2Fgvq7qbi.jpeg&w=3840&q=75)
Transcribed Image Text:**Question**
In a certain group of 63 students, 24 are taking economics, 19 are taking precalculus, and 9 are taking both. What is the probability that a randomly chosen student is taking either economics or precalculus? Give your answer as a fraction.
**Solution Explanation**
To find the probability that a student is taking either economics or precalculus, use the formula for the union of two sets.
The formula is:
\[ P(E \cup P) = P(E) + P(P) - P(E \cap P) \]
where:
- \( P(E) \) is the probability of a student taking economics.
- \( P(P) \) is the probability of a student taking precalculus.
- \( P(E \cap P) \) is the probability of a student taking both subjects.
Plugging in the numbers:
- 24 students are taking economics, so \( P(E) = \frac{24}{63} \).
- 19 students are taking precalculus, so \( P(P) = \frac{19}{63} \).
- 9 students are taking both, so \( P(E \cap P) = \frac{9}{63} \).
Now compute:
\[ P(E \cup P) = \frac{24}{63} + \frac{19}{63} - \frac{9}{63} = \frac{34}{63} \]
Therefore, the probability that a randomly chosen student is taking either economics or precalculus is \( \frac{34}{63} \).
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