Question

A certain group of symptom-free women between the ages of 40 and 50 are randomly selected to participate in mammography screening. The incidence rate of breast cancer among such women is 0.8%. The false negative rate for the mammogram is 10%. The false positive rate is 7%. If a the mammogram results for a particular woman are positive (indicating that she has breast cancer), what is the probability that she actually has breast cancer?

Answer 1

Probability of having cancer = 0.008

Probability of not having cancer = 1 - 0.008 = 0.992

Given, If woman has cancer, probability of testing negative = 0.10 (false negative)

So, If woman has cancer, probability of testing positive = 1 - 0.10 = 0.90

If woman not has cancer, probability of testing positive = 0.07 (false positive)

So, If woman not has cancer, probability of testing negative = 1 - 0.07 = 0.93

Now, using the law of conditional probability , P(B/A) = P(A and B)/P(A)   (B/A means B given A)

P (A and B) = P(A) * P(B)

P (having cancer and tests positive) = P(having cancer) * P(tests positive given she is having cancer) = 0.008 * 0.90 = 0.0072

Similiarly, P (not having cancer and tests positive) = 0.992 * 0.07 = 0.06944

P (has a positive test result) = 0.0072 + 0.06944 =  0.07664

So, P(having cancer given she has been tested positive) = P (having cancer and testing positive) / P (testing positive) = 0.0072 / 0.07664 = 0.0939

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