For independent events A and B:
AP(A ∪ B) = P(A) + P(B) - P(A)P(B); P(A ∩ B) = P(A) × P(B)
BP(A) = P(B)
CAlways exclusive
DP(A) + P(B) = 1
Answer & Solution
Correct answer: A. P(A ∪ B) = P(A) + P(B) - P(A)P(B); P(A ∩ B) = P(A) × P(B)
Independent: P(A ∩ B) = P(A) P(B). Union by inclusion-exclusion: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = P(A) + P(B) - P(A)P(B).
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