What is the probability that this happen P(A)+P(B)−2P(A ∩B)

Question

I think that my question has a bad structured, but my question is in base the next.

Let A and B be any sets. Show that the probability that exactly one of the events A or B occurs is:

$$P(A)+P(B)−2P(A\cap B)$$

I thought that this is possible with The inclusion-exclusion principle, is it right or I need other thing?

Answer 1

The inclusion exclusion principle (for any two events, $A,B$) is that:$$\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)\\~\\\text{also}\\~\\\mathsf P(A\cap B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cup B)$$

You may indeed use this to show that: $$\mathsf P((A\cup B)\cap (A\cap B)^\complement)=\mathsf P(A)+\mathsf P(B)-2\,\mathsf P(A\cap B)$$

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Also note:

$(A\cup B)\cap (A\cap B)^\complement= (A\cap B^\complement)\cup(A^\complement\cap B)$

Answer 2

The probability of the event $E:$ $A$ occurs and $B$ does not occur is $P(A)-P(A\cap B)$.

The probability of the event $F:$ $B$ occurs and $A$ does not occur is $P(B)-P(A\cap B)$.

$E$ and $F$ are mutually exclusive (disjoint), so what is $P(E\cup F)$?