Really Need Help Please

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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  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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