Prove the inclusion-exclusion formula for $P(A \cup B \cup C) $ I need to prove the following equation.
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$
I started with this..
$P(A \cup B \cup C) = P(A \cup B) \cup C$
I think that I may go on with De Morgan laws ..
but I'm not sure which are the right steps.
Can you give me some hint? Of course I'm not asking for the solution, just a "way to follow".
 A: Hint:


*

*This is a special case of probabilistic version of inclusion-exclusion principle.

*With $n=3$ (you have three sets) the most mechanic approach would be to split your sets into disjoint parts (there would be $2^3 = 8$ of these), that is,
\begin{align}
\Omega_{\varnothing} &= \{\omega \in \Omega \mid \omega \notin A, \omega \notin B, \omega \notin C\} \\
\Omega_{A} &= \{\omega \in \Omega \mid \omega \in A, \omega \notin B, \omega \notin C\} \\
\Omega_{B} &= \{\omega \in \Omega \mid \omega \notin A, \omega \in B, \omega \notin C\} \\
&\;\vdots \\
\Omega_{AB} &= \{\omega \in \Omega \mid \omega \in A, \omega \in B, \omega \notin C\} \\
&\;\vdots\\
\Omega_{ABC} &= \{\omega \in \Omega \mid \omega \in A, \omega \in B, \omega \in C\}
\end{align}
and then use $P(X \cup Y) = P(X) + P(Y)$ for disjoint $X$ and $Y$. Just expand both sides of your formula and it should match.

*A less mindless approach would be to prove it as above for $n=2$, namely $$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$$ and then apply it several times, first with $X = A \cup B$, $Y = C$, and then again for terms like $P(A \cup B)$.

*Finally, you could always try induction (i.e. prove for general $n$), however, that would require more work.


I hope this helps $\ddot\smile$
A: You simply have
$$P(A \cup B \cup C) = P((A \cup B) \cup C)
\\=P(A \cup B) +P( C)-P((A \cup B) \cap C)
\\=P(A )+P(B)-P(A \cap B) +P( C)-P((A\cap C) \cup (B \cap C)
\\ =P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
