Question

Assume that 0.1% of people from a certain population have a germ. A test gives false positive in 10% of cases when the person does not have this germ. This test gives false negative in 20% of cases when this person has this germ. Suppose you pick a random person from the population and apply this test twice. Both time it gives you positive result. What is the probability that this person actually has this germ?

Answer 1

Let D shows the event that person has germ.So we have

P(D) = 0.001, P(D') = 1 - P(D) = 1 - 0.001 = 0.999

Let P shows the event that test gives positive results and N shows the event that test gives negative results. So we have

P(P|D') = 0.10

P(N|D) = 0.20

By the complement rule we have

P(N|D') = 1 - P(P|D') = 1 - 0.10= 0.90

P(P|D) = 1 - P(N|D) = 1 - 0.20= 0.80

Let T shows the event that we get positive results is two tests. Since test are independent from each other so we have

P(T|D) = P(P|D)P(P|D) = 0.80*0.80 = 0.64

P(T|D') = P(P|D')P(P|D') = 0.10 *0.10 = 0.01

By the law of total probability, the probability that we get both test results positive is

P(T) = P(T|D)P(D) + P(T|D')P(D') = 0.64 * 0.001 + 0.01* 0.999 = 0.01063

By the Bayes theorem, the probability that this person actually has this germ given that both test give positive test results is

P(D|T) = [ P(T|D)P(D) ] / P(T) = [0.64 * 0.001] / 0.01063 = 0.0602

Answer: 0.0602