Question

In the general population, only 0.3% have colorectal cancer. (Simply, interpret this as P(Cancer)=0.003.) The hemoccult test is used in the detection of colorectal cancer as a cheaper and much less invasive alternative to colonoscopy. However it is also much less accurate. The test is positive 50% of the time if the patient has the disease, and will be positive 3% of the time if the patient does not have the disease. What is the probability of a false negative as a result of this test? Choose the closest answer.

a. 0.952

b. 0.048

c. 0.0299

d. 0.002

Answer 1

We are given here that: P( cancer ) = 0.003

Also, we are given that: P( positive | cancer ) = 0.5, therefore P( negative | cancer ) = 1 - 0.5 = 0.5

P( positive | no cancer ) = 0.03, therefore P( negative | no cancer) = 1 - 0.03 = 0.97

Using law of total probability, we get here:

P( negative ) = P( negative | cancer )P( cancer ) + P( negative | no cancer)P( no cancer )

P( negative ) = 0.5*0.003 + 0.97*(1 - 0.003) = 0.96859

Now using bayes theorem, the probability of false negative is computed here as:

P( cancer | negative ) = P( negative and cancer ) / P( negative ) = 0.5*0.003/0.96859 = 0.002

Therefore 0.002 is the required probability here.