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 d_1, 2100 has disease d₂, and others had disease d₃. 1500 patients with disease d_1, 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

Events E₁, E₂, E₃ and S be the events defined as follows:

E₁: The patient had disease d₁

E₂: The patient had disease d₂

E₃: The patient had disease d₃

S: The patient showed the symptom

E₁, E₂, and E₃ are mutually exclusive and exhaustive events.

\[P\left( E_1 \right) = \frac{1800}{5000} = \frac{18}{50}\]

\[P\left( E_2 \right) = \frac{2100}{5000} = \frac{21}{50}\]

\[P\left( E_3 \right) = \frac{1100}{5000} = \frac{11}{50}\]

Now,

\[P\left( \frac{S}{E_1} \right)\] = Probability that the patient showed symptom given that patient had disease d₁ =

\[\frac{1500}{5000} = \frac{15}{50}\]

\[P\left( \frac{S}{E_2} \right)\] = Probability that the patient showed symptom given that patient had disease d₂ = \[\frac{1200}{5000} =\frac{12}{50}\]

\[P\left( \frac{S}{E_3} \right)\]  = Probability that the patient showed symptom given that patient had disease d₃ =  \[\frac{900}{5000} = \frac{9}{50}\]

Using Bayes theorem, we have probability that patient had disease d₁ such that symptom of d₁ showed = \[P\left( \frac{E_1}{S} \right) =

\frac{P\left( E_1 \right)P\left( \frac{S}{E_1} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)} = \frac{\frac{18}{50} \times \frac{15}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{270}{621}\]

Probability that patient had disease d₂ such that symptom of d₂ showed = 

\[P\left( \frac{E_2}{S} \right) = \frac{P\left( E_2 \right)P\left( \frac{S}{E_2} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)} = \frac{\frac{21}{50} \times \frac{12}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{252}{621}\]

Probability that patient had disease d₃ such that symptom of d₃ showed = \[P\left( \frac{E_3}{S} \right) = \frac{P\left( E_3 \right)P\left( \frac{S}{E_3} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)} = \frac{\frac{11}{50} \times \frac{9}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{99}{621}\]

Thus, the patient is most likely to have the disease d₁.