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".

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') =

3 answers

Read through the nearly identical posts linked in Related Questions below. I don't know if any were answered, but it's worth looking.

And if you'd follow directions and put your SUBJECT in the SUBJECT box, that would really help.
REALLY BIG HINT. Draw the probability tree. The answers come quickly with that.
Given probabilities:
P(A) = 1/200 = 0.005
P(B) = 199/200 = 0.995
P(B|A) = 0.9
P(B|~A) = 0.1
Infer:
P(~B|A) = 0.1
P(~B|~A) = 0.9
Then:
a.) Find P(A|B)
P(A|B) = P(A)P(B|A) = 0.005×0.9 = 0.0045 rounds to 0.0
b.) Find P(~A|~B)
P(~A|~B) = P(~A)P(~B|~A) = 0.995×0.9 = 0.896 rounds to 0.9
Note: the symbol ~A means A' in your notation.
QED