Answer: Outcomes lying in both events would otherwise be counted twice in P(A) + P(B).
- A Outcomes lying in both events would otherwise be counted twice in P(A) + P(B)
- B The intersection always has probability zero and must be removed
- C Subtraction converts the union into a mutually exclusive pair of events
- D Probabilities must sum to exactly one across all events considered
Correct answer: A. Outcomes lying in both events would otherwise be counted twice in P(A) + P(B)
Explanation: P(A) and P(B) each include the shared outcomes, so the overlap is counted twice and must be removed once.
Concept context
Random experiments, sample space, events, and the axiomatic approach to probability