A doctor assumes that a patient has one of three diseasesd d

A doctor assumes that a patient has one of three diseasesd d1 d2or d3. Before any test, he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.8 if the patient has disease d1, 0.6 if he has disease d2, and 0.4 if he has disease d3. Given that the outcome of the test was positive, what probabilities should the doctor now assign to the three possible diseases?

Solution

Since initially he assumes equal probability for all the three diseases

P(d1) + P(d2) +P(d3) = 1

X + X + X=1

X = 1/3

NOw , if the patient has disease d1 , carrying out probability = 0.8

hence ,0.8 * 0.333

NOw , if the patient has disease d2 , carrying out probability = 0.6

hence ,0.6 * 0.333

NOw , if the patient has disease d3 , carrying out probability = 0.4

hence ,0.4 * 0.333

Net probability of having test positive =

0.8 * 0.33 + 0.6 * 0.33 +0.4 * 0.33 = 0.6

the test obtained positive

hence in the ratio

0.8X + 0.6X + 0.4X =1

X = 0.555

d1 = 0.8 * 0.555 = 0.4444

d2 = 0.6 * 0.555 = 0.3333

d3 = 0.4 * 0.555 =0.2222