Question

A new medical test has been designed to detect the presence of the mysterious Brainlesserian disease. Among those who have the disease, the probability that the disease will be detected by the new test is 0.7. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.02. It is estimated that 20 % of the population who take this test have the disease. If the test administered to an individual is positive, what is the probability that the person actually has the disease

Answer 1

0.8974 = 89.74% probability that the person 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: Positive test.

Event B: Has the disease.

Probability of a positive test:

0.7 of 0.2(has the disease).

0.02 of 100 - 20 = 80%(does not have the disease). Thus:

`P(A) = 0.7*0.2 + 0.02*0.8 = 0.156`

Probability of a positive test and having the disease:

0.7 of 0.2. So

`P(A \cap B) = 0.7*0.2 = 0.14`

What is the probability that the person actually has the disease?

`P(B|A) = (P(A \cap B))/(P(A)) = (0.14)/(0.156) = 0.8974`

0.8974 = 89.74% probability that the person actually has the disease.