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') =

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3. 👁 1,904
1. 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.

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Writeacher
2. REALLY BIG HINT. Draw the probability tree. The answers come quickly with that.

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3. 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

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