An incipient form of cancer occurs in three out of every 1000 Americans. To prov

Question

An incipient form of cancer occurs in three out of every 1000 Americans. To provide early detection, a screening test has been developed that rarely errs. Among healthy patients, only 5% get a + reaction (false alarm). Among patients with this incipient cancer, only 2% get a – reaction (missed alarm). If this test is used to screen the American public, all those who get a + reaction will be hospitalized for exploratory surgery. What proportion of these people who are thought to have cancer, will actually have cancer?

Explanation / Answer

Let C shows the event that person has cancer and N shows the event that person do not have cancer. So we have

P(C) = 3 /1000 = 0.003

And P(N) = 1 - 0.003 = 0.997

Now

P(+|N) = 0.05, P(-|C) = 0.02

By the complement rule we have

P(+|C) = 1 - P(+|N) = 1 - 0.05 = 0.95

By the total law of probability we have

P(+) = P(+|C)P(C) + P(+|N)P(N) = 0.95 * 0.003 + 0.05 * 0.997 = 0.0527

Now by the Baye's theorem the probability that person will actually have cancer given that he has positive test result will be

P(C|+) = [ P(+|C)P(C) ] / P(+) = [0.95* 0.003] / 0.0527 = 0.0541

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