Question

A test for detection of a particular disease is not fool proof. The test will correctly detect the disease 90% of the time, but will incorrectly detect the disease 1% of the time. For a large population of which an estimated 0.2% have the disease, a person is selected at random, given the test, and told that he has the disease. What are the chances that the person actually have the disease?

Answer 1

Let A, E₁ and E₂ denote the events that the person suffers from the disease, the test detects the disease correctly and the test does not detect the disease correctly, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{90}{100} \\ P\left(E_2\right)=\frac{1}{100} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{2}{1000} \\ P\left(A/E_2\right)=\frac{998}{1000} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{90}{100}}\times {\displaystyle \frac{2}{1000}}}{{\displaystyle \frac{90}{100}}\times {\displaystyle \frac{2}{1000}}+{\displaystyle \frac{1}{100}}\times {\displaystyle \frac{998}{1000}}} \\ =\frac{180}{180+998}=\frac{180}{1178}=\frac{90}{589} \end{gathered}$$


Disclaimer: The solution provided here is according to the question, but in the question correct and incorrect detection percentages are 90% and 1%, respectively. Their sum is 91 %. However, the ideal sum of the percentages should be 100% and the question should have been framed accordingly.