$ ext{If } P(A \cup B) = P(A \cap B) ext{ for any two events A and B, then :}$

Q: $\text{If } P(A \cup B) = P(A \cap B) \text{ for any two events A and B, then :}$

$\text{If } P(A \cup B) = P(A \cap B) \text{ for any two events A and B, then :}$

Question

$	ext{If } P(A \cup B) = P(A \cap B) 	ext{ for any two events A and B, then :}$

Accepted Answer

To understand why Option 1, "P(A) = P(B)," is the correct answer, let's analyze the given condition and the properties of probability. ### Explanation for Option 1: P(A) = P(B) The condition given is $ P(A \cup B) = P(A \cap B) $. This equation implies that the probability of either event A or event B occurring is equal to the probability of both events A and B occurring simultaneously. In probability theory, the formula for the union of two events is: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$ Given that $ P(A \cup B) = P(A \cap B) $, we can substitute this into the formula: $$ P(A \cap B) = P(A) + P(B) - P(A \cap B) $$ Rearranging the terms, we get: $$ 2P(A \cap B) = P(A) + P(B) $$ This simplifies to: $$ P(A) + P(B) = 2P(A \cap B) $$ Since $ P(A \cap B) $ is a non-negative value (as probabilities are always non-negative), the equation implies that the probabilities of A and B must be equal for the equation to hold true. Therefore, $ P(A) = P(B) $. ### Explanation for Why Other Options are Incorrect - **Option 2: P(A) > P(B)** - If $ P(A) > P(B) $, then $ P(A) + P(B) $ would not equal $ 2P(A \cap B) $ under the given condition $ P(A \cup B) = P(A \cap B) $. This would violate the equality derived from the condition, making this option incorrect. - **Option 3: P(A) < P(B)** - Similarly, if $ P(A) < P(B) $, the sum $ P(A) + P(B) $ would again not satisfy the equation $ P(A) + P(B) = 2P(A \cap B) $. This would contradict the given condition, thus making this option incorrect. - **Option 4: $ P(A) \ne P(B) $** - If $ P(A) \ne P(B) $, it implies that the probabilities of A and B are not equal, which would not satisfy the derived equation $ P(A) + P(B) = 2P(A \cap B) $. Therefore, this option is incorrect as it contradicts the necessary condition for the given equality to hold. In conclusion, the only scenario where the given condition $ P(A \cup B) = P(A \cap B) $ holds true is when $ P(A) = P(B) $, making Option 1 the correct answer.