Conditional Probability: Are my answers correct?

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 10% of the time if the person does not have the virus. (This 10% 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)= 0.89204


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') =0.991

Answer 1

P(A|B) = P(AB)/(P(B)) = P(AB)/(P(AB)) + P(A'B)) = P(B|A)P(A)/( P(B|A)P(A) +  P(B|A')P(A')) =

.9*.005/(.9*.005+.1*.995)) = .0045/(.0045+.0995)) = 45/1040 = 9/208 = .0433

 

P(A'|B') = P(A'B')/(P(B')) = P(A'B')/(P(A'B')) + P(AB')) = P(B'|A')P(A')/( P(B'|A')P(A') + P(B'|A)P(A)) =

.9*.995/(.9*.995+.1*.005) = .8955/(.8955+.0005) = .8955/.8960 = 1791/1792 = .9994