Question

Suppose it is known that 1% of the population have a certain kind of cancer. It is also known that a test for this kind of cancer is positive in 99% of the people who have it but is also positive in 2% of the people who do not have it. What is the probability that a person who tests positive has cancer of this type?

Short answer

Answer

The probability that a person who tests positive has cancer is $\frac{1}{3}$.

Step-by-step solution

Given Information

Probability of having cancer is $0.01$, the probability of positive test result is and false positive is $0.02$

Definition of Independent Event

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

Drawing the tree diagram for the situation

Draw the tree diagram depicting the person have cancer and don’t have cancer and it tested positive or negative and show respective probability.

Finding the probability that a person who tests positive has cancer

Find the probability that that a person who tests positive has cancer using the Bayes Theorem.

$$\begin{gathered} P\left(TestPositive\overline{)HasCancer}\right)=\frac{P\left(Hascancerandtestpositive\right)}{P\left(TestPositive\right)} \\ =\frac{0.99\times 0.01}{0.99\times 0.01+0.02\times 0.99} \\ =\frac{0.01}{0.01+0.02} \\ =\frac{1}{3} \end{gathered}$$