1. ## conditional probability problem

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 0.8 if the patient
has 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?

i know it involves the conditional probability formula but i can't figure out which numbers to use for each part. thanks!

2. This seems to be Bayes, but then you cannot have 2 diseases at once.
That should be a partition and I don't see it here.

Let P=psoitive and D1=Disease 1, D2=Disease 2 and D3=Disease 3.

P(D1)=P(D2)=P(D3)=1/3 our partition?

P(P)=P(P)P(P|D1)+P(D2)P(P|D2)+P(D3)P(P|D3)=[.8+.6+.4]/3=.6

Thus P(D1|P)=P(D1P)/P(P)=(.8)(1/3)/.6=4/9.