Question

Let E and F be two events such that \( P(E) > 0 \) and \( P(F) > 0 \). Which one of the following is NOT equivalent to the condition that \( P(E) = P(E \cap F) \)?

• A. \( P(F) = P(F \cap E) \)
• B. E and F are independent
• C. \( E^c \) and F are independent
• D. \( P(E^c) P(F^c) \neq P(E^c \cap F^c) \)

Hint:

When analyzing conditional probability and independence, carefully check how events interact and whether they are independent or dependent based on the conditions.

Answer 1

The Correct Option is D

The condition \( P(E) = P(E \cap F) \) indicates that event \( E \) is fully contained within event \( F \). This condition can be equivalent to the following:

Step 1: Analyze the options.
- (a) \( P(F) = P(F \cap E) \): This is true because if \( P(E) = P(E \cap F) \), it implies \( E \) is contained in \( F \), so the probability of \( F \) is the same as the probability of the intersection of \( F \) and \( E \). - (b) E and F are independent: If \( P(E) = P(E \cap F) \), this suggests that \( E \) and \( F \) are dependent, not independent. This is not necessarily true in this case. - (c) \( E^c \) and F are independent: The independence of \( E^c \) and \( F \) follows from the condition \( P(E) = P(E \cap F) \), and this holds true in such cases. - (d) \( P(E^c) P(F^c) \neq P(E^c \cap F^c) \): This statement is the correct answer because the condition \( P(E) = P(E \cap F) \) implies \( E \) and \( F \) are dependent. However, this does not directly imply that \( P(E^c) P(F^c) \) is not equal to \( P(E^c \cap F^c) \), which contradicts the equivalence. Thus, the correct answer is (d).