Conditional Probability

A certain virus infects one in every 400 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. 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.

(b) Find the probability that a person does not have the virus given that they have tested negative.

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  1. Given:
    P(A)=1/400
    P(B|A)=9/10
    P(B|~A)=1/10

    By the law of complements,
    P(~A)=1-P(A)=399/400
    By the law of total probability,
    P(B)=P(B|A)*P(A)+P(B|A)*P(~A)
    =(9/10)*(1/400)+(1/10)*(399/400)
    =51/500
    Note: get used to working in fraction when doing probability.

    (a) Find P(A|B):
    By Baye's Theorem,
    P(A|B)
    =P(B|A)*P(A)/P(B)
    =(9/10)*(1/400)/(51/500)
    =3/136

    (b) Find P(~A|~B)
    We know that
    P(~A)=1-P(A)=399/400
    P(~B)=1-P(B)=133/136
    P(A∩B)
    =P(B|A)*P(A) [def. of cond. prob.]
    =9/10*(1/400)
    =9/4000

    P(A∪B)
    =P(A)+P(B)-P(A∩B)
    =1/400+51/500-9/4000
    =409/4000

    P(~A|~B)
    =P(~A∩~B)/P(~B)
    =P(~A∪B)/P(~B)
    =(1-P(A∪B)/(1-P(B)) [ law of complements ]
    =(3591/4000) ÷ (449/500)
    =3591/3592

    The results can be easily verified using a contingency table for a random sample of 4000 persons (assuming outcomes correspond exactly to probability):
    ===....B...~B...TOT
    ..A . 9 . . 1 . . 10
    .~A .399 .3591 . 3990
    Tot .408 .3592 . 4000

    So P(A|B)=9/408=3/136
    P(~A|~B)=3591/3592
    As before.

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  2. Note:
    The numbers in the contingency table can also be obtained by a tree diagram
    considering a sample of 4000 people.

    400--A-->10--B-->9
    -------------~B->1
    ----~A-->3990--B-->399
    --------------~B-->3591

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  3. Probability of 1/300 8/80 1/8

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