probability of complementary

probability of complementary

For any event $A,P\left(A\right)+P\left(A^{\prime }\right)=1$ , from which $P\left(A\right)=1-P\left(A^{\prime }\right)$.

\begin{proof} In Axiom 3, let $k=2$, $A_1=A$ , and $A_2=A^{\prime }$. Since by definition of $A^{\prime }$ , $A\cup A^{\prime }=\mathcal{S}$ while $A$ and $A^{\prime }$ are disjoint, $1=P\left(\mathcal{S}\right)=P\left(A\cup A^{\prime }\right)=P\left(A\right)+P\left(A^{\prime }\right)$. \end{proof}

This proposition is surprisingly useful because there are many situations in which $P\left(A^{\prime }\right)$ is more easily obtained by direct methods than is $P\left(A\right)$ .

EX 2.13 components in series