Question

Probability of event A is 0.7 and event B is 0.4, and $ P(A \cap B^c) = 0.5 $, then the value of $ P(B | A \cup B^c) $ is equal to:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{3}{4} \)

Hint:

In conditional probability problems, break down the events and use the laws of probability to simplify the calculations.

Answer 1

The Correct Option is C

We are given the following: - \( P(A) = 0.7 \) - \( P(B) = 0.4 \) - \( P(A \cap B^c) = 0.5 \) We need to find \( P(B | A \cup B^c) \).
The formula for conditional probability is: \[ P(B | A \cup B^c) = \frac{P(B \cap (A \cup B^c))}{P(A \cup B^c)} \] Now, calculate \( P(A \cup B^c) \): \[ P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) \] Since \( P(B^c) = 1 - P(B) = 1 - 0.4 = 0.6 \), and \( P(A \cap B^c) = 0.5 \), we have: \[ P(A \cup B^c) = 0.7 + 0.6 - 0.5 = 0.8 \] Next, calculate \( P(B \cap (A \cup B^c)) \).
This can be broken down as: \[ P(B \cap (A \cup B^c)) = P(B \cap A) + P(B \cap B^c) = P(B \cap A) \quad (\text{since } P(B \cap B^c) = 0) \] We use the relationship: \[ P(B) = P(B \cap A) + P(B \cap A^c) \] Therefore: \[ P(B \cap A) = P(B) - P(B \cap A^c) = 0.4 - (P(A) - P(A \cap B^c)) = 0.4 - (0.7 - 0.5) = 0.4 - 0.2 = 0.2 \] Thus: \[ P(B \cap (A \cup B^c)) = 0.2 \] Finally: \[ P(B | A \cup B^c) = \frac{0.2}{0.8} = \frac{1}{4} \]