If A and B are mutually exclusive, then $P(A \cup B) = $:
A$P(A) + P(B) + P(A \cap B)$
B$P(A) + P(B)$
C$P(A) - P(B)$
D$P(A) \cdot P(B)$
Answer & Solution
Correct answer: B. $P(A) + P(B)$
Mutually exclusive: $A \cap B = \phi$, so $P(A \cap B) = 0$. Thus $P(A \cup B) = P(A) + P(B)$.
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