Question

A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".

(a) Using Bayes’ Theorem, when a person tests positive, determine the probability that the person is infected.
(b) Using Bayes’ Theorem, when a person tests negative, determine the probability that the person is not infected.

Answer 1

A) P(A|B) = 0.01966

B) P(A'|B') = 0.99944

Explanation:

A) We are told that A is the event "the person is infected" and B is the event "the person tests positive".

Thus, using bayes theorem, the probability that the person is infected is; P(A|B)

From bayes theorem,

P(A|B) = [P(A) × P(B|A)]/[(P(A) x P(B|A)) + (P(A') x P(B|A'))]

Now, from the question,

P(A) = 1/400

P(A') = 399/400

P(B|A) = 0.8

P(B|A') = 0.1

Thus;

P(A|B) = [(1/400) × 0.8)]/[((1/400) x 0.8) + ((399/400) x (0.1))]

P(A|B) = 0.01966

B) we want to find the probability that when a person tests negative, the person is not infected. This is;

P(A'|B') = P(Not infected|negative) = P(not infected and negative) / P(negative) = [(399/400) × 0.9)]/[((399/400) x 0.9) + ((1/400) x (0.2))] = 0.99944