MHB Probability that exactly one of three events will occur

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The discussion focuses on deriving the probability that exactly one of three events, A1, A2, and A3, occurs. The formula presented is Pr(A1) + Pr(A2) + Pr(A3) - 2Pr(A1 ∩ A2) - 2Pr(A2 ∩ A3) - 2Pr(A1 ∩ A3) + 3Pr(A1 ∩ A2 ∩ A3). The user attempts to break down the calculation by expressing the required probability in terms of intersections and complements of the events. They suggest using the principle of inclusion-exclusion to simplify the notation and derive the probability step by step. The conversation emphasizes the mathematical approach to solving the problem systematically.
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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

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

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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Alexmahone said:
Let $A_1$, $A_2$, and $A_3$ be three arbitrary events. Show that the probability that exactly one of these three events will occuris
$\Pr(A_1)+\Pr(A_2)+\Pr(A_3)-2\Pr(A_1\cap A_2)-2\Pr(A_2\cap A_3)-2\Pr(A_1\cap A_3)+3\Pr(A_1\cap A_2\cap A_3)$
To simplify notation.
$P(AB^cC^c)=P(AB^c)-P(AB^cC)$
$=P(A)-P(AB)-[P(AC)-P(ABC)]$
$=P(A)-P(AB)-P(AC)+P(ABC)$

Do that twice more for $P(A^cBC^c)~\&~P(A^cB^cC)$ and add.
 
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