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#### Alexmahone

##### Active member

- Jan 26, 2012

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Let $A_1$, $A_2$, and $A_3$ be three arbitrary events. Show that the probability that exactly one of these three events will occur is

$\Pr(A_1)+\Pr(A_2)+\Pr(A_3)-2\Pr(A_1\cap A_2)-2\Pr(A_2\cap A_3)-2\Pr(A_1\cap A_3)+3\Pr(A_1\cap A_2\cap A_3)$

The required probability$=\Pr(A_1\cap A_2^c\cap A_3^c)+\Pr(A_1^c\cap A_2\cap A_3^c)+\Pr(A_1^c\cap A_2^c\cap A_3)$

$\Pr(A_1\cap A_2^c\cap A_3^c)=\Pr(A_1\cap(A_2\cup A_3)^c)$

$=\Pr(A_1)-\Pr(A_1\cap(A_2\cup A_3))$

How do I proceed?

$\Pr(A_1)+\Pr(A_2)+\Pr(A_3)-2\Pr(A_1\cap A_2)-2\Pr(A_2\cap A_3)-2\Pr(A_1\cap A_3)+3\Pr(A_1\cap A_2\cap A_3)$

__My attempt:__The required probability$=\Pr(A_1\cap A_2^c\cap A_3^c)+\Pr(A_1^c\cap A_2\cap A_3^c)+\Pr(A_1^c\cap A_2^c\cap A_3)$

$\Pr(A_1\cap A_2^c\cap A_3^c)=\Pr(A_1\cap(A_2\cup A_3)^c)$

$=\Pr(A_1)-\Pr(A_1\cap(A_2\cup A_3))$

How do I proceed?

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