Among 500 freshmen pursuing a business degree at a university, 338 are enrolled in an economics course, 220 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in each of the following? (Enter your answers to three decimal places.) (a) an economics and/or a mathematics course (b) exactly one of these two courses (c) neither an economics course nor a mathematics course

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**Probability and Set Theory in University Course Enrollment**

Let's explore a practical application of probability and set theory among university students.

**Problem Statement:**
Among 500 freshmen pursuing a business degree at a university, 338 are enrolled in an economics course, 220 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in each of the following? (Enter your answers to three decimal places):

(a) An economics and/or a mathematics course

(b) Exactly one of these two courses

(c) Neither an economics course nor a mathematics course

**Solution Approach:**

To solve these problems, we will use the principle of inclusion and exclusion for sets and basic probability formulas.

**(a) Probability of Enrolled in an Economics and/or a Mathematics Course:**

The probability of being enrolled in an economics and/or a mathematics course is given by:
\[ P(E \cup M) = P(E) + P(M) - P(E \cap M) \]

Where:
- \( P(E) \) is the probability of being enrolled in an economics course
- \( P(M) \) is the probability of being enrolled in a mathematics course
- \( P(E \cap M) \) is the probability of being enrolled in both courses

Given:
- \( P(E) = \frac{338}{500} \)
- \( P(M) = \frac{220}{500} \)
- \( P(E \cap M) = \frac{123}{500} \)

Thus,
\[ P(E \cup M) = \frac{338}{500} + \frac{220}{500} - \frac{123}{500} \]
\[ P(E \cup M) = \frac{338 + 220 - 123}{500} \]
\[ P(E \cup M) = \frac{435}{500} \]
\[ P(E \cup M) = 0.870\]

**(b) Probability of Enrolled in Exactly One of the Two Courses:**

The probability of being enrolled in exactly one of the two courses is given by:
\[ P(\text{Exactly one}) = P(E \setminus M) + P(M \setminus E) \]

Where:
- \( P(E \setminus M) \) is the probability of being enrolled in only an economics course
-
Transcribed Image Text:**Probability and Set Theory in University Course Enrollment** Let's explore a practical application of probability and set theory among university students. **Problem Statement:** Among 500 freshmen pursuing a business degree at a university, 338 are enrolled in an economics course, 220 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in each of the following? (Enter your answers to three decimal places): (a) An economics and/or a mathematics course (b) Exactly one of these two courses (c) Neither an economics course nor a mathematics course **Solution Approach:** To solve these problems, we will use the principle of inclusion and exclusion for sets and basic probability formulas. **(a) Probability of Enrolled in an Economics and/or a Mathematics Course:** The probability of being enrolled in an economics and/or a mathematics course is given by: \[ P(E \cup M) = P(E) + P(M) - P(E \cap M) \] Where: - \( P(E) \) is the probability of being enrolled in an economics course - \( P(M) \) is the probability of being enrolled in a mathematics course - \( P(E \cap M) \) is the probability of being enrolled in both courses Given: - \( P(E) = \frac{338}{500} \) - \( P(M) = \frac{220}{500} \) - \( P(E \cap M) = \frac{123}{500} \) Thus, \[ P(E \cup M) = \frac{338}{500} + \frac{220}{500} - \frac{123}{500} \] \[ P(E \cup M) = \frac{338 + 220 - 123}{500} \] \[ P(E \cup M) = \frac{435}{500} \] \[ P(E \cup M) = 0.870\] **(b) Probability of Enrolled in Exactly One of the Two Courses:** The probability of being enrolled in exactly one of the two courses is given by: \[ P(\text{Exactly one}) = P(E \setminus M) + P(M \setminus E) \] Where: - \( P(E \setminus M) \) is the probability of being enrolled in only an economics course -
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