Question 1047554: "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 d2, and 0.4 if he has d3. Given that the outcome of the test was positive, what probabilities should the doctor now assign to the three possible diseases?"
My try on the question: I assumed that an equal probability meant 1/3. There are three diseases, so if the patient has one disease, the others are excluded: does this mean there is another factor in the calculation?
Otherwise, I concluded that 0.8*(1/3) for d1, 0.6*(1/3) for d2, and 0.4*(1/3) for d3.
But it seems to be too easy among the questions I have been solving.
Thank you in advance.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Given that the outcome of the test was positive,
what probabilities should the doctor now assign to the three possible diseases?"
That Given points to using Bayes Theorem
P(A|B) = P(A and B)/P(B)
P(d1 | pos)
etc for D2 and D3
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