Question

If \(A\) and \(B\) are two such events that \(P(A \cup B) = P(A \cap B)\), then which of the following is true?

• A. \(P(A) + P(B) = 0\)
• B. \(P(A) + P(B) = P(A)P(B|A)\)
• C. \(P(A) + P(B) = 2P(A)P(B|A)\)
• D. None of the above

Hint:

Always convert \(P(A \cap B)\) into conditional form when options involve \(P(B|A)\).

Answer 1

The Correct Option is C

To determine which statement is true given that \(P(A \cup B) = P(A \cap B)\), let's analyze the problem step by step:

1. Using the formula for the probability of the union of two events \(A\) and \(B\):
   • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
2. Given the condition \(P(A \cup B) = P(A \cap B)\), substitute this into the formula:
   • \(P(A \cap B) = P(A) + P(B) - P(A \cap B)\).
3. By rearranging this equation, we get:
   • \(2P(A \cap B) = P(A) + P(B)\).
4. We know that the probability of the intersection of two events can be expressed in terms of conditional probability:
   • \(P(A \cap B) = P(A)P(B|A)\).
5. Substitute \(P(A \cap B)\) from step 4 into the equation from step 3:
   • \(2P(A)P(B|A) = P(A) + P(B)\).
6. Hence, the true statement based on the given condition is:
   • \(P(A) + P(B) = 2P(A)P(B|A)\).

Therefore, the correct answer is \(P(A) + P(B) = 2P(A)P(B|A)\).