A doctor assumes that a patient has one of three diseases d1, d2, or d3. Before any test, he assumes an equal probability for each disease. He carries out a test that will be positive with probability .8 if the patient has d1, .6 if he has disease d2, and .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:
Assuming equal probability for all the three diseases
P(d1) + P(d2) +P(d3) = 1
X + X + X=1
3X = 1 X = 1/3
Applying Baye’s formula:
\[P(pos) = ( P( pos| d1) .d1 + P( pos| d2) .d2 + P( pos| d3) .d3 )\]
If the patient has disease d1, probability = 0.8
= 0.8 * 0.33
If the patient has disease d2 , probability = 0.6
= 0.6 * 0.33
If the patient has disease d3 , probability = 0.4
= 0.4 * 0.33
Probability of having test positive =
0.8 * 0.33 + 0.6 * 0.33 +0.4 * 0.33 = 0.6
the test obtained positive
0.8X + 0.6X + 0.4X =1
x = 1/1.8
X = 0.555
d1 = 0.8 * 0.555 = 0.4444
d2 = 0.6 * 0.555 = 0.3333
d3 = 0.4 * 0.555 =0.2222