# Principle of inclusion-exclusion proof

1. Sep 16, 2011

### mottov2

There are 3 events A,B and C prove that

P(A$\cup$B$\cup$C) = P(A)+P(B)+P(C)-P(A$\cap$B)-P(A$\cap$C)-P(B$\cap$C)+P(A$\cap$B$\cap$C)
each event is disjoint so by the additivity rule....

My attempt:
A$\cup$B$\cup$C = (A$\cap$B$\cap$C)$\cup$(A$\cap$[A$\cap$B$\cap$C]c)$\cup$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)$\cup$(C$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c)$\cup$(B$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c)

each event is disjoint so by the additivity rule....
P(A$\cup$B$\cup$C) = P(A$\cap$B$\cap$C)+P(A$\cap$[A$\cap$B$\cap$C]c)+P([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)+P(C$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c)+P(B$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c)

P(A$\cap$(A$\cap$B$\cap$C)c) = P(A)-P(A$\cap$B$\cap$C)
P((B$\cap$C)$\cap$(A$\cap$B$\cap$C)c) = P(B$\cap$C)-P(A$\cap$B$\cap$C)
P(C$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c) = P(C$\cap$(A$\cap$C)c$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c) = P(C)-P(A$\cap$C)-P((B$\cap$C)$\cap$(A$\cap$B$\cap$C)c) = P(C)-P(A$\cap$C)-P(B$\cap$C)+P(A$\cap$B$\cap$C)
P(B$\cap$Ac$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c) = P(B$\cap$(A$\cap$B)c$\cap$([B$\cap$C]$\cap$[A$\cap$B$\cap$C]c)c) = P(B)-P(A$\cap$B)-P((B$\cap$C)$\cap$(A$\cap$B$\cap$C)c) = P(B)-P(A$\cap$B)-P(B$\cap$C)+P(A$\cap$B$\cap$C)

then by substitution...

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

did i do this right? I feel like i may have overcomplicated it..

Last edited: Sep 16, 2011
2. Sep 16, 2011

### Stephen Tashi

You haven't explained what laws of probability can be assumed in order to prove the result and you didn't say which laws you used. Putting a 'P' in front of each set in an expression is not "substitution".

3. Oct 9, 2011

### M-quest

Proof By Induction is a lot more elegant,but you have to be careful of your notation