Question

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 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". Hint: Make a Tree Diagram 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') = % Get help: Video

Answer 1

ANSWER :

A certain virusbinfects one in every 200 people.Atest used to detect the virus in a person is positiuve85%of the time if the person has the virus and 5% of the time if the person doesnot have the virus.

p(infection) = 1/200 = P(A)

P(A') = 199 / 200

P(test is positive / person has the virus) = 0.85

Pr (B/A) = 0.85

P(test is positive / person donot have the virus) = 0.05

P(B/A') = 0.05

P(B'/A') = 0.95

(a)

P(Person is infected / test positive) =

(B)

P( person is not infected / tested negative)=

  

.