3. Of 500 newly admitted college students, 220 are enrolled in College Algebra, 320 are enrolled in English 1, and 125 are enrolled in both. Part A: Determine the probability that the college student is enrolled in College Algebra. Part B: Determine the probability that the college student is enrolled in both courses. Part C: Determine the probability that the student is enrolled in either College Algebra or English 1 but not both.
3. Of 500 newly admitted college students, 220 are enrolled in College Algebra, 320 are enrolled in English 1, and 125 are enrolled in both. Part A: Determine the probability that the college student is enrolled in College Algebra. Part B: Determine the probability that the college student is enrolled in both courses. Part C: Determine the probability that the student is enrolled in either College Algebra or English 1 but not both.
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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Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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![### Probability Analysis of Course Enrollment
**Problem Statement:**
Of 500 newly admitted college students, 220 are enrolled in College Algebra, 320 are enrolled in English 1, and 125 are enrolled in both.
**Part A**: Determine the probability that the college student is enrolled in College Algebra.
**Part B**: Determine the probability that the college student is enrolled in both courses.
**Part C**: Determine the probability that the student is enrolled in either College Algebra or English 1 but not both.
---
### Solution
#### Part A: Probability of Enrollment in College Algebra
To find the probability that a randomly selected student is enrolled in College Algebra, we use the formula for probability:
\[ P(\text{College Algebra}) = \frac{\text{Number of students enrolled in College Algebra}}{\text{Total number of students}} \]
Here, the number of students enrolled in College Algebra is 220.
\[ P(\text{College Algebra}) = \frac{220}{500} = 0.44 \]
Thus, the probability that a college student is enrolled in College Algebra is **0.44** or **44%**.
#### Part B: Probability of Enrollment in Both Courses
To find the probability that a student is enrolled in both College Algebra and English 1:
\[ P(\text{Both Courses}) = \frac{125}{500} = 0.25 \]
Thus, the probability that a college student is enrolled in both courses is **0.25** or **25%**.
#### Part C: Probability of Enrollment in Either College Algebra or English 1, but not Both
To determine the probability that a student is enrolled in either College Algebra or English 1 but not both, we can use the formula:
\[ P(\text{Either, Not Both}) = P(\text{College Algebra Only}) + P(\text{English 1 Only}) \]
First, we need to find the number of students enrolled only in College Algebra and only in English 1.
- Students enrolled only in College Algebra: \( 220 - 125 = 95 \)
- Students enrolled only in English 1: \( 320 - 125 = 195 \)
Now, we find the probabilities:
\[ P(\text{College Algebra Only}) = \frac{95}{500} \approx 0.19 \]
\[ P(\text{English 1 Only}) = \frac{195](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5ed07d5-7a8f-4097-9c81-82df457ca627%2F5cb8f7a1-fd2c-43c2-ab65-0a089e80ca23%2Fipm94cg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Analysis of Course Enrollment
**Problem Statement:**
Of 500 newly admitted college students, 220 are enrolled in College Algebra, 320 are enrolled in English 1, and 125 are enrolled in both.
**Part A**: Determine the probability that the college student is enrolled in College Algebra.
**Part B**: Determine the probability that the college student is enrolled in both courses.
**Part C**: Determine the probability that the student is enrolled in either College Algebra or English 1 but not both.
---
### Solution
#### Part A: Probability of Enrollment in College Algebra
To find the probability that a randomly selected student is enrolled in College Algebra, we use the formula for probability:
\[ P(\text{College Algebra}) = \frac{\text{Number of students enrolled in College Algebra}}{\text{Total number of students}} \]
Here, the number of students enrolled in College Algebra is 220.
\[ P(\text{College Algebra}) = \frac{220}{500} = 0.44 \]
Thus, the probability that a college student is enrolled in College Algebra is **0.44** or **44%**.
#### Part B: Probability of Enrollment in Both Courses
To find the probability that a student is enrolled in both College Algebra and English 1:
\[ P(\text{Both Courses}) = \frac{125}{500} = 0.25 \]
Thus, the probability that a college student is enrolled in both courses is **0.25** or **25%**.
#### Part C: Probability of Enrollment in Either College Algebra or English 1, but not Both
To determine the probability that a student is enrolled in either College Algebra or English 1 but not both, we can use the formula:
\[ P(\text{Either, Not Both}) = P(\text{College Algebra Only}) + P(\text{English 1 Only}) \]
First, we need to find the number of students enrolled only in College Algebra and only in English 1.
- Students enrolled only in College Algebra: \( 220 - 125 = 95 \)
- Students enrolled only in English 1: \( 320 - 125 = 195 \)
Now, we find the probabilities:
\[ P(\text{College Algebra Only}) = \frac{95}{500} \approx 0.19 \]
\[ P(\text{English 1 Only}) = \frac{195
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