A genetic test used to determine whether a patient has a specific type of cancer

Question

A genetic test used to determine whether a patient has a specific type of cancer. The probability of a test coming back positive when cancer is not present (called a false-positive) is 0.03, where as the probability of a test coming back positive when cancer is present is 0.92.

If the frequency of cancer in the population is 15%, what is the probability a person is cancer free if the test is positive?

Answer = ___ %

Note: Your answer should be typed in as a percentage. Include two decimal places in your response.

Explanation / Answer

P(cancer is present) = 0.15

P(test is positive | cancer is not present) = 0.03

P(test is positive | cancer is present) = 0.92

P(test is positive) = P(test is positive | cancer is not present) * P(cancer is not present) + P(test is positive | cancer is present) * P(cancer is present)

                            = 0.03 * (1 - 0.15) + 0.92 * 0.15

                            = 0.1635

P(cancer is not present | test is positive) = P(test is positive | cancer is not present) * P(cancer is not present) / P(test is positive)

                                                                 = 0.03 * (1 - 0.15) / 0.1635

                                                                 = 0.1560

                                                                 = 15.6% (ans)

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