Can someone help explain this question? Please show all your work. The proportio

Question

Can someone help explain this question? Please show all your work.

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). Let C = a randomly selected man has prostate cancer Use 3 decimal places. P(C | Pos) = 0.398 P(C^c | Pos) =

Explanation / Answer

Let's use the letters And B to denote the following:

A : number of people who have a cancer and tested positive

B : number of people who don't have a cancer but still tested positive

From the table given above, we see:

A = 69

B = 11

So,

P(C | Pos) = A / (A+B) = 69/(69+11) = 0.8625

P(Cc | Pos) = 1 - P(C | Pos) = 0.1375

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