# Prove that? : P(AuuBuuC)=P(A)+P(B)+P(C)-P(AnnB)-P(BnnC)-P(AnnC)+P(AnnBnnC)

## $P \left(A \cup B \cup C\right) = P \left(A\right) + P \left(B\right) + P \left(C\right) - P \left(A \cap B\right) - P \left(B \cap C\right) - P \left(A \cap C\right) + P \left(A \cap B \cap C\right)$ I can show that with diagram, but how to prove it with formulas?

Aug 7, 2018

#### Explanation:

$\text{Prerequisite : } P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) \ldots . \left(\star\right)$.

$P \left(A \cup B \cup C\right) = P \left(A \cup D\right) , \text{ where, } D = B \cup C$,

$= P \left(A\right) + P \left(D\right) - P \left(A \cap D\right) \ldots \ldots \ldots . \left[\because , \left(\star\right)\right]$,

$= P \left(A\right) + \textcolor{red}{P \left(B \cup C\right)} - \textcolor{b l u e}{P \left[A \cap \left(B \cup C\right)\right]}$,

$= P \left(A\right) + \textcolor{red}{P \left(B\right) + P \left(C\right) - P \left(B \cap C\right)} - \textcolor{b l u e}{P \left(A \cap B\right) \cup \left(A \cap C\right)} ,$

=P(A)+P(B)+P(C)-P(BnnC)-color(blue){[P(AnnB)+P(AnnC)-P((AnnB)nn(AnnC)],

$= P \left(A\right) + P \left(B\right) + P \left(C\right) - P \left(A \cap B\right) - P \left(B \cap C\right) - P \left(A \cap C\right) + P \left(A \cap B \cap C\right) ,$

as desired!