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For any two events $A$ and $B$, the probability that exactly one of them occurs is
A$P(A)+P(B)-2P(A\cap B)$
B$P(A\cap B)$
C$P(A\cup B)+P(A\cap B)$
D$1-P(A\cup B)$
Answer & Solution
Correct answer: A. $P(A)+P(B)-2P(A\cap B)$
Exactly one of $A$ or $B$ means either $A$ occurs without $B$, or $B$ occurs without $A$. So
$$P(\text{exactly one})=P(A\cup B)-P(A\cap B).$$
Using $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ gives $P(A)+P(B)-2P(A\cap B)$.
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