Question

A girl is concerned she might have a disease that affects 1% of the population. Luckily, a drug-store test is available. The test has a false-positive rate of 3% and a false-negative rate of 1% (a false positive is when a person doesnt have the disease, but the test result is positive).

(a) What is the probability of a positive test result?

(b) The girl takes the test, and it returns positive. Given this information, what is the probability that she has the disease?

(c) The girl is aware that the test is imperfect, so she decides to take a second test. It also returns positive. Given the results of both tests, what is the probability that she has the disease? You may assume that the two test results are conditionally independent given her disease status.

Answer 1

Here, we are given that: P( disease ) = 0.01

Also, we are given that: P( positive | no disease ) = 0.03, and
P( negative | disease ) = 0.01

a) P( positive ) = P( positive | no disease ) P( no disease ) + P( positive | disease ) P( disease )

P( positive ) = 0.03*0.99 + 0.99*0.01 = 0.0396

Therefore P( positive ) = 0.0396

b) Using bayes theorem, the probability here is computed as:

P( disease | positive ) = P( positive | disease ) P( disease ) / P( positive )

P( disease | positive ) = 0.99*0.01 / 0.0396 = 0.25

Therefore 0.25 is the required probability here.

c) The probabilities here are computed as:

P( both positive | no disease ) = 0.03*0.03 = 0.0009

P( both positive | disease ) = 0.99*0.99 = 0.9801

Using law of total probability, we get here:

P( both positive ) = P( both positive | no disease ) P( no disease ) + P( both positive | disease ) P( disease )

P( both positive ) = 0.0009*0.99 + 0.9801*0.01 = 0.010692

Therefore using bayes theorem now, we get here:

P( disease | both positive ) =  P( both positive | disease ) / P( both positive )

P( disease | both positive ) =  0.9801*0.01 / 0.010692 = 0.9167

Therefore 0.9167 is the required probability here.