Question

# A certain virus infects one in every 200 people. A test used to detect the virus...

A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 85% 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(A|B). Round your answer to the nearest tenth of a percent.
P(A|B)=_________%

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.
P(A'|B') = ________%

A: The person has virus

B: Tested positive

P(A) = 1/200 = 0.005

P(A') = 0.995

a) P(A | B) = P(A & B)/P(B) [Bayes' Theorem]

P(A and B) = 0.005x0.85 = 0.00425

P(A' and B) = 0.995x0.08 = 0.0796

P(a person has the virus given that they have tested positive) = 0.00425/(0.00425 + 0.0796)

= 0.051

= 5.1%

b) P(A' | B') = P(A' and B') / P(B')

P(A' and B') = 0.995x0.92 = 0.9154

P(A and B') = 0.005x0.15 = 0.00075

P(a person does not have the virus given that they test negative) = 0.9154/(0.9154+0.00075)

= 0.999

= 99.9%

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