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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?
Solution
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.
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