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If $A$ and $B$ are mutually exclusive events, then the addition theorem reduces to
A$P(A\cup B)=P(A)P(B)$
B$P(A\cup B)=P(A)+P(B)$
C$P(A\cup B)=P(A)+P(B)+P(A\cap B)$
D$P(A\cup B)=1-P(A\cap B)$
Answer & Solution
Correct answer: B. $P(A\cup B)=P(A)+P(B)$
In general, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. For mutually exclusive events, $A\cap B=\varnothing$, so $P(A\cap B)=0$. Hence $P(A\cup B)=P(A)+P(B)$.
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