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

Answer:

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

Step-by-step explanation:

Conditional Probability

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

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.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.00552}{0.03534} = 0.1562$

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