A doctor assumes that patient has one of three diseases d 1, ~d 2 or d 3. Before any test he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.7 if the patient has disease d 1,0.5 if the patient has disease d2 and 0.8 if the patient has disease d3. Given that the outcome of the test was positive then probability that patient has disease d2 is

Step-by-step solution

The patient has an equal prior probability of having each disease, so P(D₁) = P(D₂) = P(D₃) = 1 3 . The conditional probabilities of a positive test are P(T|D₁) = 0.7 , P(T|D₂) = 0.5 , and P(T|D₃) = 0.8 . The total probability of a positive test is calculated using the law of total probability: P(T) = P(T|D₁)P(D₁) + P(T|D₂)P(D₂) + P(T|D₃)P(D₃) = 1 3 (0.7 + 0.5 + 0.8) = 1 3 (2.0) = 2 3 . Applying Bayes' theorem gives: P(D₂|T) = P(T|D₂)P(D₂) P(T) = 0.5 1 3 2 3 = 0.5 2 = 1 4 . Final answer: 1 4 , which corresponds to