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My book says that for any two events A and B

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

The proof it provides is this:

$$\def\P{\mathop{\rm P}}\begin{align}\P(A \cup B) &= \P(A \setminus B) + \P(A \cap B) + \P(B \setminus A) \\[1ex]&= \P(A \setminus B) + \P(A \cap B) + \P(B \setminus A) + \P(A\cap B) - \P(A\cap B) \\[1ex]&= \P(A)\hspace{16.5ex}+\P(B)\hspace{16.5ex}-\P(A\cap B) \end{align}$$

My question is what happened to the $P(A\setminus B)$ and $P(B\setminus A)$ in the second line?

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1 Answer 1

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$P(A\setminus B)+P(A\cap B)=P(A)$ since these are disjoint events, whose union is $A$. Likewise for $B$.

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  • $\begingroup$ Oh I see. Drawing a picture I see what you have wrote. Thank you! $\endgroup$
    – user865043
    Dec 22, 2020 at 5:50

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