Question

In certain parts of the world tuberculosis (a very treatable and curable illness) is present in 25% of the people. If a simple chest X-ray can be used as a diagnostic tool with an accuracy of 99% detection of TB if it is present. Only in 0.5% of the cases, normal people get a positive diagnosis. 2 If a person is selected at random and the diagnosis is positive, what is the probability that the person is actually infected.

Answer 1

0.9851 = 98.51% probability that the person is actually infected.

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: Infected

Probability of a positive test:

99% of 25%(People have the disease).

0.5% of 100 - 25 = 75%(People do not have the disease). So

`P(A) = 0.99*0.25 + 0.005*0.75 = 0.25125`

Probability of a positive test and being infected:

99% of 25%, so:

`P(A \cap B) = 0.99*0.25 = 0.2475`

What is the probability that the person is actually infected?

`P(B|A) = (P(A \cap B))/(P(A)) = (0.2475)/(0.25125) = 0.9851`

0.9851 = 98.51% probability that the person is actually infected.