Refer to Example 2.26. Assume that the proportion of

Step 1 of 3

Given that, the proportion of people in the community who have the disease is 0.05.

Let D denote the event that a person has the disease.

Let  +ve represent the event that the test is positive. The values are given by

\(\begin{aligned}
P(D)=0.05, P\left(D^{c}\right) & =1-P(D) \\
& =1-0.05 \\
& =0.95 \\
P(+v e / D)=0.99, \text { and } P(+ & v e / D c)=1-P(+v e / D) \\
& =1-0.99 \\
& =0.01 .
\end{aligned}\)