The Law of Total Probability

The Law of Total Probability

Let $A_1,\dots ,A_k$ be mutually exclusive and exhaustive events. Then for any other event $B$ ,

$$\begin{array}{rl}P\left(B\right)& =P\left(B\mid A_1\right)P\left(A_1\right)+\cdots +P\left(B\mid A_k\right)P\left(A_k\right)\\ & =\sum _{i=1}^{k}P\left(B\mid A_i\right)P\left(A_i\right)\end{array}\text{\hspace{1em}(2.5)}$$

\begin{proof}

• Because the ${\mathrm{A}}_i$‘s are mutually exclusive and exhaustive,
  • then if $B$ occurs, it must be in conjunction with exactly one of the ${\mathrm{A}}_i$‘s.
• That is, $B=\left(A_1\cap B\right)\cup \dots \cup \left(A_k\cap B\right)$ ,
  • where the events $\left(A_i\cap B\right)$ are mutually exclusive.
• This “partitioning of $B$ ” is illustrated in Figure 2.11.

Figure 2.11 Partition of $B$ by mutually exclusive and exhaustive $A_i$‘s

Thus

$$P\left(B\right)=\sum _{i=1}^{k}P\left(A_i\cap B\right)=\sum _{i=1}^{k}P\left(B\mid A_i\right)P\left(A_i\right)$$

as desired. \end{proof}