Homework help in probability please?

Question

A doctor assumes that a patient has one of three disease a1,a2, or a3. 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 a1, 0.6 if he has disease a2, and 0.4 if he has disease a3. Given that the outcome of the test was positive, what probabilities should the doctor now assign to the three possible diseases?

If one can explain the answer, I greatly appreciate it.

Answer 1

This is bayesian analysis.

p(a1) = p(a2) = p(a3) = 1/3
p(positive | a1) = 0.8
p(positive | a2) = 0.6
p(positive | a2) = 0.4

Find p(a3 | positive)

p(a3 | pos) = p(a3 ∩ pos) / p(pos)
= p(a1)*p(pos | a1)
/{p(a1)*p(pos | a1) + p(a2)*p(pos | a2) + p(a3)*p(pos | a3)}
= (1/3)(.8) / {(1/3)(.8 + .6 + .4}
= 8 / (8 + 6 + 4) = 8/18 = 4/9

Answer 2

Did you understand northstar ?
It was good. I added a little explanation.

This is bayesian analysis.

p(a1) = p(a2) = p(a3) = 1/3
p(positive | a1) = 0.8
p(positive | a2) = 0.6
p(positive | a3) = 0.4

Find p(a3 | positive)

p(a3 | pos)
= p(a3 ∩ pos) / p(pos)
= p(pos | a3)* p(a3) / p(pos)
here p(a3, pos) is p(pos|a3) x p(a3)
= 0.4 * (1/3) / p(pos)

now, what is the p(pos). It is the total prob of conditional prob (meaning sum up all the possible ways positive result can occurs)
p(pos) = p(positive | a1)*p(a1) + p(positive | a2)*p(a2) +p(positive | a3)*p(a3) = 1/3(1.8) = 0.6

Answer 3

all diseases have 1/3 probability to start, bc there are 3 of them.
now just multiply 1/3 times probability of each teset
a1=.26666666666(repeating)=4/15
a2=.2=1/5
a3=1.3333333333(repeating)=2/15