For independent events A and B:
- P(A ∪ B) = P(A) + P(B) - P(A)P(B); P(A ∩ B) = P(A) × P(B)
- P(A) = P(B)
- Always exclusive
- P(A) + P(B) = 1
Show answer and 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).