Question

Let \( A \) and \( B \) be two events such that \( P(A \cup B) = P(A) + P(B) - P(A)P(B) \). If \( 0 < P(A) < 1 \) and \( 0 < P(B) < 1 \), then \( P(A \cup B)' = \)

• A. \( 1 - P(A) \)
• B. \( 1 - P(A') \)
• C. \( 1 - P(A)P(B) \)
• D. \( [1 - P(A)]P(B') \)
• E. \( 1 \)

Hint:

For independent events, the probability that neither occurs is the product of the probabilities that each individual event does not occur: \( P(A \cup B)' = P(A')P(B') \).

Answer 1

The Correct Option is D

Concept: The Addition Rule for any two events states \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Comparing this to the given equation, we find that \( P(A \cap B) = P(A)P(B) \), which implies that events \( A \) and \( B \) are independent.

Step 1: Determine the relationship between events.
Given: \( P(A \cup B) = P(A) + P(B) - P(A)P(B) \). From the general addition rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Equating the two, we get \( P(A \cap B) = P(A)P(B) \). This confirms \( A \) and \( B \) are independent events.

Step 2: Express the complement of the union.
We need to find \( P(A \cup B)' \), which is \( 1 - P(A \cup B) \). \[ P(A \cup B)' = 1 - [P(A) + P(B) - P(A)P(B)] \] \[ P(A \cup B)' = 1 - P(A) - P(B) + P(A)P(B) \]

Step 3: Factorize the expression.
Group the terms: \[ P(A \cup B)' = (1 - P(A)) - P(B)(1 - P(A)) \] \[ P(A \cup B)' = (1 - P(A))(1 - P(B)) \] Since \( 1 - P(A) = P(A') \) and \( 1 - P(B) = P(B') \), we can also write this as: \[ P(A \cup B)' = [1 - P(A)]P(B') \]