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.

$$\begin{gathered} \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} \\ \mathrm{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} \\ \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)+P\left(E_3\right)P\left(A/E_3\right)} \\ =\frac{{\displaystyle \frac{1800}{5000}}\times {\displaystyle \frac{1500}{1800}}}{{\displaystyle \frac{1800}{5000}}\times {\displaystyle \frac{1500}{1800}}+{\displaystyle \frac{2100}{5000}}\times {\displaystyle \frac{1200}{2100}+\frac{1100}{5000}\times \frac{900}{1100}}} \\ =\frac{15}{15+12+9}=\frac{15}{36}=\frac{5}{12} \\ \mathrm{Required}\mathrm{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{{\displaystyle \frac{2100}{5000}}\times {\displaystyle \frac{1200}{2100}}}{{\displaystyle \frac{1800}{5000}}\times {\displaystyle \frac{1500}{1800}}+{\displaystyle \frac{2100}{5000}}\times {\displaystyle \frac{1200}{2100}+\frac{1100}{5000}\times \frac{900}{1100}}} \\ =\frac{12}{15+12+9}=\frac{12}{36}=\frac{1}{3} \\ \mathrm{Required}\mathrm{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{{\displaystyle \frac{1100}{5000}}\times {\displaystyle \frac{900}{1100}}}{{\displaystyle \frac{1800}{5000}}\times {\displaystyle \frac{1500}{1800}}+{\displaystyle \frac{2100}{5000}}\times {\displaystyle \frac{1200}{2100}+\frac{1100}{5000}\times \frac{900}{1100}}} \\ =\frac{9}{15+12+9}=\frac{9}{36}=\frac{1}{4} \end{gathered}$$

As P(E₁/A ) is maximum, so it is most likely that the person suffers from the disease d₁.