Question

The proportion of people in a given community who have a certain disease is 0.005. A test

is available to diagnose the disease. If a person has the disease, the probability that the test

will produce a positive signal is 0.99. If a person does not have the disease, the probability

that the test will produce a positive signal is 0.01. If a person tests positive, what is the

probability that the person actually has the disease?

Answer 1

Solution:

Consider N be the persons which have true diseases and + sign indicates the positive sign of the test. And consider $N^{\prime}$ be the persons which have not true diseases and + sign indicates the positive sign of the test. So, the conditional probability that if a person tests positive (X) given actually disease is calculated as:

$\begin{gathered} P(X \mid+)=\frac{P(X \cap+)}{P(+)} \\ =\frac{P(+X) P(X)}{P(+X) P(X)+P\left(+X^{\prime}\right) P\left(X^{\prime}\right)} \\ =\frac{0.99 \times 0.005}{(0.99 \times 0.005)+(0.01 \times(1-0.005))} \\ =0.3322 \end{gathered}$

The required conditional probability that if a person test positive given actual disease is approximately 0.3322 .