Question

A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 8% of the time if the person does not have the virus. (This 8% 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) Find the probability that a person has the virus given that they have tested positive, i.e. find P(AB). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(AB)= | % b) Find the probability that a person does not have the virus given that they test negative, i.e. find P(A'B'). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(A'B') = License Points possible: 1 Unlimited attempts. Score on last attempt: (0,0). Score in gradebook: (0,0), Out of: (0.5, 0.5)

Answer 1

P(A) = 1/200 = 0.005

P(B | A) = 0.9

P(B | A') = 0.08

a) P(B) = P(B | A) * P(A) + P(B | A') * P(A') = 0.9 * 0.005 + 0.08 * (1 - 0.005) = 0.0841

P(A | B) = P(B | A) * P(A) / P(B) = 0.9 * 0.005 / 0.0841 = 0.054 = 5.4%

b) P(B') = 1 - P(B) = 1 - 0.0841 = 0.9159

P(B' | A') = 1 - P(B | A') = 1 - 0.08 = 0.92

P(A' | B') = P(B' | A') * P(A') / P(B') = 0.92 * (1 - 0.005) / 0.9159 = 0.999 = 99.9%