Chapter 2, Problem 143SE

Show that Theorem 2.7 holds for conditional probabilities. That is, if P(B) > 0, then $P(A|B)=1-P(\bar{A}|B)$.

To determine

Show that $P\left(A|B\right)=1-P\left(\overline{A}|B\right)$ if $P\left(B\right)>0$.

Explanation of Solution

Calculation:

From the given information, $P\left(B\right)>0$.

Let $B=\left(B\cap A\right)\cup \left(B\cap \overline{A}\right)$.

The probability of Event B is obtained as follows:

$\begin{array}{c}P\left(B\right)=P\left(\left(B\cap A\right)\cup \left(B\cap \overline{A}\right)\right)\\ =P\left(B\cap A\right)+P\left(B\cap \overline{A}\right)\langle \begin{array}{l}\text{Using addition rule of probability}\\ \text{P}\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\end{array}\rangle \\ 1=\frac{P\left(B\cap A\right)}{P\left(B\right)}+\frac{P\left(B\cap \overline{A}\right)}{P\left(B\right)}\\ 1=P\left(A|B\right)+P\left(\overline{A}|B\right)\langle \begin{array}{l}\text{Using conditional rule}\\ P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\end{array}\rangle \\ P\left(A|B\right)=1-P\left(\overline{A}|B\right)\end{array}$

Hence, the proof.