U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find: a. P(UnV) b. P(U\V) PILLY)
U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find: a. P(UnV) b. P(U\V) PILLY)
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.CR: Chapter 12 Review
Problem 83CR
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![## Problem: Mutually Exclusive Events
Given:
- Events \( U \) and \( V \) are mutually exclusive.
- \( P(U) = 0.26 \)
- \( P(V) = 0.37 \)
Find the following probabilities:
### a. \( P(U \cap V) \)
### b. \( P(U \cup V) \)
### c. \( P(U | V) \)
**Solution:**
**a. Finding \( P(U \cap V) \)**
Since \( U \) and \( V \) are mutually exclusive, their intersection is empty. Therefore,
\[ P(U \cap V) = 0 \]
**b. Finding \( P(U \cup V) \)**
The Union of \( U \) and \( V \) for mutually exclusive events:
\[ P(U \cup V) = P(U) + P(V) \]
\[ P(U \cup V) = 0.26 + 0.37 \]
\[ P(U \cup V) = 0.63 \]
**c. Finding \( P(U | V) \)**
For mutually exclusive events, the conditional probability \( P(U | V) \) is given by:
\[ P(U | V) = \frac{P(U \cap V)}{P(V)} \]
Since \( P(U \cap V) = 0 \) (from part a),
\[ P(U | V) = \frac{0}{0.37} = 0 \]
So, the answers are:
- \( P(U \cap V) = 0 \)
- \( P(U \cup V) = 0.63 \)
- \( P(U | V) = 0 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faed2933e-b67f-42a0-9fac-b6dce0e828b0%2F944e7068-b64c-48fc-a9a3-c6886187d5d3%2Fvk9zw8d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem: Mutually Exclusive Events
Given:
- Events \( U \) and \( V \) are mutually exclusive.
- \( P(U) = 0.26 \)
- \( P(V) = 0.37 \)
Find the following probabilities:
### a. \( P(U \cap V) \)
### b. \( P(U \cup V) \)
### c. \( P(U | V) \)
**Solution:**
**a. Finding \( P(U \cap V) \)**
Since \( U \) and \( V \) are mutually exclusive, their intersection is empty. Therefore,
\[ P(U \cap V) = 0 \]
**b. Finding \( P(U \cup V) \)**
The Union of \( U \) and \( V \) for mutually exclusive events:
\[ P(U \cup V) = P(U) + P(V) \]
\[ P(U \cup V) = 0.26 + 0.37 \]
\[ P(U \cup V) = 0.63 \]
**c. Finding \( P(U | V) \)**
For mutually exclusive events, the conditional probability \( P(U | V) \) is given by:
\[ P(U | V) = \frac{P(U \cap V)}{P(V)} \]
Since \( P(U \cap V) = 0 \) (from part a),
\[ P(U | V) = \frac{0}{0.37} = 0 \]
So, the answers are:
- \( P(U \cap V) = 0 \)
- \( P(U \cup V) = 0.63 \)
- \( P(U | V) = 0 \)
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