Problem 2) If P(A | B) > P(A), prove that P(A' | B) < P(A').
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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![### Problem 2)
**Given:**
If \( P(A \mid B) > P(A) \),
**Prove:**
\( P(A' \mid B) < P(A') \).
**Explanation:**
This probability problem involves conditional probabilities \( P(A \mid B) \) and \( P(A' \mid B) \) in relation to unconditional probabilities \( P(A) \) and \( P(A') \).
Here:
- \( P(A \mid B) \) is the probability of event \(A\) occurring given that event \(B\) has occurred.
- \( P(A) \) is the unconditional probability of event \(A\) occurring.
- \( P(A' \mid B) \) is the probability of the complement of event \(A\) (denoted as \( A' \), meaning \( A \) does not occur) given that event \( B \) has occurred.
- \( P(A') \) is the unconditional probability of event \(A\) not occurring.
The goal is to demonstrate that if the probability of \(A\) given \(B\) is greater than the unconditional probability of \(A\), this influences the relationship of the complement probabilities such that \( P(A' \mid B) < P(A') \).
### Solution Approach
Using the properties of conditional probabilities and the relationship between complementary events can aid in solving this problem. Here is the detailed breakdown of the process:
1. **Relative Complements**: Remember that \( A \) and \( A' \) are relative complements, which means \( P(A) + P(A') = 1 \). Similarly, \( P(A \mid B) + P(A' \mid B) = 1 \).
2. **Given Condition**: Start with the given condition \( P(A \mid B) > P(A) \).
3. **Use Complements in Conditional Probability**: Using the given condition, transform the problem into another statement involving the complement.
\[
P(A' \mid B) = 1 - P(A \mid B)
\]
\[
P(A') = 1 - P(A)
\]
4. **Substitute the Given Condition**: Substitute the given condition \( P(A \mid B) > P(A) \) into the complement expressions.
\[
P(A' \mid B) = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4fcf4b0b-d82c-43e8-86bb-da9864dd2a82%2F9f2b75ae-5da0-403c-8c04-827d07a4a017%2F5lwvfmt_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 2)
**Given:**
If \( P(A \mid B) > P(A) \),
**Prove:**
\( P(A' \mid B) < P(A') \).
**Explanation:**
This probability problem involves conditional probabilities \( P(A \mid B) \) and \( P(A' \mid B) \) in relation to unconditional probabilities \( P(A) \) and \( P(A') \).
Here:
- \( P(A \mid B) \) is the probability of event \(A\) occurring given that event \(B\) has occurred.
- \( P(A) \) is the unconditional probability of event \(A\) occurring.
- \( P(A' \mid B) \) is the probability of the complement of event \(A\) (denoted as \( A' \), meaning \( A \) does not occur) given that event \( B \) has occurred.
- \( P(A') \) is the unconditional probability of event \(A\) not occurring.
The goal is to demonstrate that if the probability of \(A\) given \(B\) is greater than the unconditional probability of \(A\), this influences the relationship of the complement probabilities such that \( P(A' \mid B) < P(A') \).
### Solution Approach
Using the properties of conditional probabilities and the relationship between complementary events can aid in solving this problem. Here is the detailed breakdown of the process:
1. **Relative Complements**: Remember that \( A \) and \( A' \) are relative complements, which means \( P(A) + P(A') = 1 \). Similarly, \( P(A \mid B) + P(A' \mid B) = 1 \).
2. **Given Condition**: Start with the given condition \( P(A \mid B) > P(A) \).
3. **Use Complements in Conditional Probability**: Using the given condition, transform the problem into another statement involving the complement.
\[
P(A' \mid B) = 1 - P(A \mid B)
\]
\[
P(A') = 1 - P(A)
\]
4. **Substitute the Given Condition**: Substitute the given condition \( P(A \mid B) > P(A) \) into the complement expressions.
\[
P(A' \mid B) = 1
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