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 $P(A)+P(B)-2P(A\cap B)$.
Related questions
The number of positive integral solutions of $x_1+x_2+\cdots+x_r=n$ isFor two events $A$ and $B$, which formula correctly gives the probability of $A\cup B$?
!The probability that a marksman will hit a target is given as $\frac{1}{5}$. Then the probThe number of ways in which 10 persons can go in two boats, so that there may be 5 on eachThe mean deviation from the median isIf events $A$ and $B$ satisfy $P(B)\neq 0$, which expression correctly defines the conditiWhich property of conditional probability is correct?If $P(A)=0.5$ and $P(B/A)=0.4$, then by the multiplication theorem, $P(A\cap B)$ equals