The proportion of men in the population who have prostate cancer is 35 per 100,0

Question

The proportion of men in the population who have prostate cancer is 35 per 100,000. In 1980, a new test was developed to test for the presence of prostate cancer. The researchers who developed the test wanted to know the probability that a man who tested positive for prostate cancer actually had the disease. To find this probability, they took a random sample of 99 men known to have prostate cancer and another random sample of 150 men known not to have prostate cancer (after having a biopsy). Their test is a radioimmunoassay for prostate acid phosphatase (RIA-PAP). They found the following results (real data). TABLE ABOVE.



Let C = a randomly selected man has prostate cancer
Use 3 decimal places.

(a) P(C | Pos) = ??       

(b) P(Cc | Pos) = ???     

NOTE: (a) is NOT 0.862 nor 0.863 nor 0.697
            (b) is NOT 0.138 nor 0.303
Please explain if possible. Thanks in advance!

Explanation / Answer

(a) P(C| Pos) = P(C and Pos)/P(Pos) = 69/80 = 0.863

(b) P(Cc | Pos) = P(Cc and Pos)/P(Pos) = 11/80 = 0.138

(The above answers appear to be correct. Why do you say these are not the correct answers?)

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