A certain virus infects one in every 400 people. A test used to detect the virus

Question

A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 5% of the time if the person does not have the virus. (This 5% 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 and do not include a percent sign.
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 and do not include a percent sign.
P(A'|B') = %

Explanation / Answer

a)P(B)=P(A)*(B|A)+P(Ac)*P(B|Ac)=(1/400)*(0.8)+(399/400)*0.05=0.051875

hence P(A|B)=(1/400)*(0.8)/0.051875=0.038554 ~3.9

b)P(B')=1-P(B)=1-0.051875 =0.948125

P(A'|B') =(399/400)*0.95/0.948125 =0.999473 ~99.9

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