Question

A certain disease has an incidence rate of 0.6%. If the false negative rate is 8% and the false positive rate is 3%, compute the probability that a person who tests positive actually has the disease.

Answer 1

0.1562 = 15.62% probability that a person who tests positive actually has the disease.

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Tests positive

Event B: Has the disease.

Probability of a positive test:

100 - 8 = 92% of 0.6%(person has the disease).

3% of 100 - 0.6% = 99.40%(person does not have the disease). So

`P(A) = 0.92*0.006 + 0.03*0.994 = 0.03534`

Probability of testing positive and having the disease:

92% of 0.6%. So

`P(A \cap B) = 0.92*0.006 = 0.00552`

Probability that a person who tests positive actually has the disease.

`P(B|A) = (P(A \cap B))/(P(A)) = (0.00552)/(0.03534) = 0.1562`

0.1562 = 15.62% probability that a person who tests positive actually has the disease.