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

$P(A\cup B)$ 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 $$P(A\cup B)-P(A\cap B)$$


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

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


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

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

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

Tags:

Probability

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