Confusion with regards to the phrase "exactly one of the events occurs"

Suppose we have events $A$ and $B$. We want to write the probability that exactly one of the events $A,B$ occurs in terms of $P(A),P(B)$ and $P(A \cap B)$ only

My thought: Since I want only one occurring, $A$ or $B$, we must find $P(A \cup B)$ which equals $P(A) + P(B) - P(A \cap B)$.. However, on my answer sheet it says the answer is $P(A) + P(B) - 2P(A \cap B )$. Am I missing something?

Solution 1:

$P(A\cup B)$ includes those outcomes where both events $A$ and $B$ occur. We want to exclude this case, because if they both occur it is not exactly one occurring. This undesirable case is exactly $P(A\cap B)$, so we need to subtract it, as $$P(A\cup B)-P(A\cap B)$$

Solution 2:

As vadim123 mentions, $A\cup B$ includes the outcome where $A$ and $B$ happens so we must subtract these off. We have

$$\begin{align*} P(A\cap B^C)+P(A^C \cap B) &=P(A\cup B)-P(A\cap B)\\\\ &=\left(P(A)+P(B)-P(A \cap B)\right)-P(A\cap B)\\\\ &=P(A)+P(B)-2P(A \cap B) \end{align*}$$

Solution 3:

What you are looking for is $$P(A\cup B)-P(A\cap B)=$$

$$ P(A)+P(B)-P(A\cap B)-P(A\cap B)=$$

$$P(A)+P(B)-2P(A\cap B)$$