Question:
Let \( E \) and \( F \) be two events with \( 0<P(E)<1 \), \( 0<P(F)<1 \) and \( P(E) + P(F) \geq 1 \). Which of the following statements is (are) TRUE?
Let \( E \) and \( F \) be two events with \( 0<P(E)<1 \), \( 0<P(F)<1 \) and \( P(E) + P(F) \geq 1 \). Which of the following statements is (are) TRUE?
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When dealing with conditional probabilities, recall that \( P(E|F) = \frac{P(E \cap F)}{P(F)} \), and use the relationships between events and their complements.
Updated On: Nov 20, 2025
- \( P(E \cap F) \leq P(E) \)
- \( P(E \cup F)<P(E^C \cup F^C) \)
- \( P(E|F^C) \geq P(F^C|E) \)
- \( P(E^C | F) \leq P(F | E^C) \)
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The Correct Option is A, C, D
Solution and Explanation
Step 1: Analyze the options.
We are given conditions on the probabilities of two events \( E \) and \( F \), and we need to evaluate the truth of the given statements. - (A) \( P(E \cap F) \leq P(E) \): This is true because \( P(E \cap F) \leq P(E) \) by the definition of conditional probability.
- (B) \( P(E \cup F)<P(E^C \cup F^C) \): This is false because \( P(E \cup F) \geq P(E^C \cup F^C) \) based on the inclusion-exclusion principle.
- (C) \( P(E|F^C) \geq P(F^C|E) \): This is true because \( P(E|F^C) \geq P(F^C|E) \) follows from the relationships between conditional probabilities.
- (D) \( P(E^C | F) \leq P(F | E^C) \): This is true because conditional probabilities relate in such a way that \( P(E^C | F) \leq P(F | E^C) \) holds.
Step 2: Conclusion.
The correct answers are (A), (C), and (D).
We are given conditions on the probabilities of two events \( E \) and \( F \), and we need to evaluate the truth of the given statements. - (A) \( P(E \cap F) \leq P(E) \): This is true because \( P(E \cap F) \leq P(E) \) by the definition of conditional probability.
- (B) \( P(E \cup F)<P(E^C \cup F^C) \): This is false because \( P(E \cup F) \geq P(E^C \cup F^C) \) based on the inclusion-exclusion principle.
- (C) \( P(E|F^C) \geq P(F^C|E) \): This is true because \( P(E|F^C) \geq P(F^C|E) \) follows from the relationships between conditional probabilities.
- (D) \( P(E^C | F) \leq P(F | E^C) \): This is true because conditional probabilities relate in such a way that \( P(E^C | F) \leq P(F | E^C) \) holds.
Step 2: Conclusion.
The correct answers are (A), (C), and (D).
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