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?

Answer 1

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.