Addition Theorem for Two Events

The addition theorem states that for any two events $A$ and $B$ of a sample space $S$, $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ This was already listed as property 7 in the previous section, but because it is used so heavily it deserves its own proof, given here two ways.

Proof by decomposition. Any two events $A$ and $B$ let us split $A\cup B$ into two pieces that share nothing in common: the part of $A$ outside $B$, namely $A\cap B'$, and $B$ itself. That is, $A\cup B=(A\cap B')\cup B$, and these two pieces are mutually exclusive. By property 10 (probabilities add over mutually exclusive events), $P(A\cup B)=P(A\cap B')+P(B)$. By property 8, $P(A\cap B')=P(A)-P(A\cap B)$. Substituting gives $P(A\cup B)=P(A)-P(A\cap B)+P(B)=P(A)+P(B)-P(A\cap B)$.

Proof by Venn diagram (Fig. 9.1). Let $n(S)=n$ be the total outcomes, $n(A)=x$, $n(B)=y$, and $n(A\cap B)=z$ be the outcomes common to both. Drawing $A$ and $B$ as overlapping circles inside the rectangle $S$, the overlap region holds $z$ outcomes, the part of $A$ outside $B$ holds $x-z$, and the part of $B$ outside $A$ holds $y-z$. Adding the three non-overlapping regions, $n(A\cup B)=(x-z)+z+(y-z)=x+y-z$, i.e. $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Dividing every term by $n(S)$ turns outcome counts into probabilities and gives the same result: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. …