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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?

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  • $\begingroup$ Imagine the sample space. Suppose you flip two coins, $A$ is "at least one head" and $B$ is "at least one tail". $\endgroup$
    – Ethan Bolker
    Apr 3, 2018 at 0:21
  • $\begingroup$ @Jimmy: Just use **asterisks** for bold text, you don't need to abuse TeX math formatting for this. $\endgroup$
    – user856
    Apr 3, 2018 at 11:41

3 Answers 3

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

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  • $\begingroup$ @Henry Yes, that's what Vadim means: $P(A\cup B)$ includes those outcomes where both events $A$ and $B$ occur (as well as those events where only one of $A$ or $B$ occurs). He just didn't mention the part in parentheses, since the OP already seemed to be aware of it. $\endgroup$
    – LarsH
    Apr 3, 2018 at 16:17
  • $\begingroup$ @LarsH - I had missed includes $\endgroup$
    – Henry
    Apr 3, 2018 at 17:45
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As vadim123 mentions, $A\cup B$ includes the outcome where $A$ and $B$ happens so we must subtract these off. We have

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

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  • $\begingroup$ Is $\cdot^C$ really the conventional way of writing the complement of an event? That's a horrible convention, in particular when the events are named $A$, $B$ etc.. (Though by no means uniquely horrible; I'm looking at you, $B = QTZ^T$...) Why not write it $P(A\cap\not B)$ or $P(A\cap\neg B)$? $\endgroup$
    – leftaroundabout
    Apr 3, 2018 at 11:15
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    $\begingroup$ Well $\lnot B$ is sometimes used, but a distinct set operation is much preferred rather than overloading the logical operation. So various conventions use: $B^\mathrm c, B', \overline B, B^\complement$, and such to indicate complementation. @leftaroundabout $\endgroup$
    – Graham Kemp
    Apr 3, 2018 at 11:51
  • $\begingroup$ @Henry Your comment is correct and doesn't contradict anything in this answer. $\endgroup$
    – Mark S.
    Apr 3, 2018 at 17:36
  • $\begingroup$ @MarkS. - I had missed includes $\endgroup$
    – Henry
    Apr 3, 2018 at 17:45
  • $\begingroup$ @leftaroundabout For the complement I never found a widespread notation that I liked. For that specific case, though, I'd use $P(A \setminus B) + P(B \setminus A)$, which is standard. (However, the "intersection&complement" variant notation can be more readable in some contexts, I think.) $\endgroup$
    – chi
    Apr 3, 2018 at 18:42
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What you are looking for is $$P(A\cup B)-P(A\cap B)=$$

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

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

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