SOLUTION: 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 5% of the time if the perso

Question 1048696: 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 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.
What I have : p=(the person tests positive =90% *(1/200)=+0.05*199/200)=0.05425=0.1
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.
What I have : p(the person with positive test is really infected)=[90%*(1/200)] divided (0.05425)=0.082949=82.94

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
What you originally have for part (a), delete it. What you have for part (b), put it under part (a). That will make it correct, by Bayes' Rule and the rule of total probability.

(b)
P(A'|B') =P(A'∩B')/P(B') = P((A∪B)')/(1-P(B)) = (1-P(A∪B))/(1-P(B))
= (1-P(A)-P(B) +P(A∩B))/(1-P(B))
= (nearest tenth of a percent.)
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