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.

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

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.