addition rule of probability

When events $A$ and $B$ are mutually exclusive, $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right).$ For events that are not mutually exclusive, adding $P\left(A\right)$ and $P\left(B\right)$ results in “double-counting” outcomes in the intersection. The next result, the addition rule for a double union probability, shows how to correct for this.

Property

For any two events $A$ and $B$ ,

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

\begin{proof} Note first that $A\cup B$ can be decomposed into two disjoint events, $A$ and $B\cap A^{\prime }$; the latter is the part of $B$ that lies outside $A$ (see Figure 2.4).

Figure 2.4 Representing $A\cup B$ as a union of disjoint events

Furthermore, $B$ itself is the union of the two disjoint events $A\cap B$ and $A^{\prime }\cap B$ , so P\left( B\right) =$ $P\left( {A \cap B}\right) + P\left( {{A}' \cap B}\right) . Thus

$$\begin{array}{rl}P\left(A\cup B\right)& =P\left(A\right)+P\left(B\cap A^{\prime }\right)\\ & =P\left(A\right)+\left[P\left(B\right)-P\left(A\cap B\right)\right]\\ & =P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\end{array}$$

\end{proof}

EX 2.14 services of households