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Question
Let d1, d2, d3 be three mutually exclusive diseases. Let S be the set of observable symptoms of these diseases. A doctor has the following information from a random sample of 5000 patients: 1800 had disease d1, 2100 has disease d2 and the others had disease d3. 1500 patients with disease d1, 1200 patients with disease d2 and 900 patients with disease d3 showed the symptom. Which of the diseases is the patient most likely to have?
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Solution
Let A, E1, E2 and E3 denote the events that the patient shows symptoms S, has disease d1, has disease d2 and has disease d3, respectively.
\[\therefore P\left( E_1 \right) = \frac{1800}{5000}\]
\[ P\left( E_2 \right) = \frac{2100}{5000} \]
\[ P\left( E_3 \right) = \frac{1100}{5000}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{1500}{1800}\]
\[P\left( A/ E_2 \right) = \frac{1200}{2100}\]
\[P\left( A/ E_3 \right) = \frac{9000}{1100}\]
\[\text{ Using Bayes' theorem, we get }\]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{1800}{5000} \times \frac{1500}{1800}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}\]
\[ = \frac{15}{15 + 12 + 9} = \frac{15}{36} = \frac{5}{12}\]
\[\text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_2 \right)P\left( A/ E_2 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{2100}{5000} \times \frac{1200}{2100}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}\]
\[ = \frac{12}{15 + 12 + 9} = \frac{12}{36} = \frac{1}{3}\]
\[\text{ Required probability }= P\left( E_3 /A \right) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{1100}{5000} \times \frac{900}{1100}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}\]
\[ = \frac{9}{15 + 12 + 9} = \frac{9}{36} = \frac{1}{4}\]
As P(E1/A ) is maximum, so it is most likely that the person suffers from the disease d1.
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