Question

A certain test for mononucleosis has a 99% chance of correctly diagnosing a patient with mononucleosis and a 5% Chance of miss diagnosing a patient who does not have the infection. Suppose the test is given to a group where 1% of the people have mononucleosis. If a randomly selected patients test result is positive, What is the probability that she has mononucleosis?

Answer 1

Let M be the event that a person has mono, and D the event that the test gives a positive result (regardless of whether M occurs at the same time).

We're told that P(D | M) = 0.99 and P(D | M') = 0.05, and for a given group of people, P(M) = 0.01. (Note: M' denotes the complement of M.)

We want to find P(M | D).

By the definition of condition probability,

P(M | D) = P(M and D)/P(D)

By the law of total probability,

P(D) = P(D and M) + P(D and M')

and using the definition of conditional probability we can rewrite this as

P(D) = P(D | M) P(M) + P(D | M') P(M')

Putting everything together gives Bayes' rule,

P(M | D) = [P(D | M) P(M)] / [P(D | M) P(M) + P(D | M') P(M')]

Plug in everything we know:

P(M | D) = (0.99 * 0.01) / (0.99 * 0.01 + 0.05 * (1 - 0.01)) ≈ 0.17