Welcome to our community

Be a part of something great, join today!

Probability that exactly one of three events will occur

Alexmahone

Active member
Jan 26, 2012
268
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?
 
Last edited:

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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.