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?

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