Question

A new, non-invasive colon cancer screening method boasts a sensitivity of 99%. That is, given that a patient has colon cancer, the screening method has a 0.99 probability of yielding a

positive test. The test is also 90% specific, meaning that if a person without colon cancer is screened, there is a 0.9 probability of a negative test result. Among the population of adults

over 45 years of age, the proportion who have colon cancer is 0.0013 (thirteen out of every ten-thousand).

Let + denote having a positive test result, - as having a negative test result, C denotes having

colon cancer, and C^c as not having colon cancer. *note the / means given not divide. Also, Bayes theorem should be used here

1a. For the population of adults over 45 years old, find the following probabilities: P(C), P(C^c), P(+/C), P(-/C), P(-/C^c), P(+/C^c).

1b. Find the overall probability of getting a positive test result for a person randomly selected from the population of adults over 45 years old, P(+)

1c. A person is randomly selected from the population of adults over 45 years old and screened for colon cancer. The test result is positive. They might be tempted to think this means they are 99% likely to have colon cancer. Uses Bayes theorem to find the actual probability that they have colon cancer, given their positive test result. P(C/+)

1d. How does the answer of part C change if we only screened people in the population who we thought have a .5 probability of colon cancer? That is, repeat the problem using P(C)=0.5

Answer 1

Solution:

1a. Given that P (C) = 0.0013, P (+/C) = 0.99, P (-/C^c) = 0.90

We know that P (C) + P (C^c) = 1

0.0013 + P (C^c) = 1

P (C^c) = 1- 0 .0013 = 0.9987

Similarly, P (+/C^c) = 1 - P (+/C) = 1 - 0.99 = 0.01

Similarly, P (-/C) = 1 - P (-/C^c) = 1 - 0.90 = 0.10

1b. P (+) = P (C)*P (+/C) + P (C^c)* P (+/C^c)

P (+) = 0.0013*0.99 + 0.9987*0.01

P (+) = 0.0113

1c. P (C/+) = [P (C)*P (+/C)]/P (+)

P (C/+) = [0.0013*0.99]/0.0113

P (C/+) = 0.1142

1d. When P (C) = 0.5

P (+) = P (C)*P (+/C) + P (C^c)* P (+/C^c)

P (+) = 0.5*0.99 + 0.5*0.01

P (+) = 0.5

P (C/+) = [P (C)*P (+/C)]/P (+)

P (C/+) = [0.5*0.99]/0.5

P (C/+) = 0.99

The probability increases to 0.99.