D1: .50
D2: .50
Suppose also that medical research has established the probability associated with each of three symptoms (denoted S1, S2, and S3) that may accompany the two diseases. That is, suppose that, given diseases D1 and D2, the probabilities that the patient will have symptoms S1, S2, or S3 are as follows: P(S1│D1) = .25; P(S2│D1) = .15; P(S3│D1) = .65; P(S1│D2) = .10; P(S2│D2) = .15; P(S3│D2) = .20.
I followed the Bayes theory and this is the answers I got. Of course they are wrong.
b. The patient has symptom S2. D1 0.04 and D2 0.06 (3 pts)
c. The patient has symptom S3. D1 0.833 and D2 0.166 (3 pts)
d. For a patient with symptom S1 in part (a), we also find symptom S3. The posterior probabilities are: D1 0.4839 and D2 0.5161 (3 pts)
Can you please help me
Oh ok. Well that makes quite a bit more sense. Remember what Bayes says (paraphrased and in non-legalese):
If I have a collection of events, we can call them A1, A2, A3. . .An that cover an entire space, and further we have a collection of sub-events, B1, B2, B3. . .Bn - then the P(of B occurring) = P(A1)P(B|A1)+. . .P(An)P(B|An).
How about an example: Say I know (I don't know) that in the United States, 54% of the population are females and 46% of the population are males. I further know that among females, 2% have cancer, and among males it is 3%. How would I go about finding the probability that someone has cancer in the United States?
Looking at Bayes I know my exhaustive events (the events that cover the entire sample space) are P(M) and P(F). I also know that the sub-event is the P(Cancer|Male) as well as for females. Using Bayes Theorem, I can find out what P(Cancer) is. Try it.