CAN A TUTOR HELP ME WITH THIS PROBLEM PLEASE!

A certain virus infects one in every 600 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. To understand the solution given below, you will need to know a few definitions or identities:

    1. P(A)=probability of event A will happen.

    2. definition of conditional probability:
    P(A|B)=P(A∧B)/P(B)
    i.e. Probability of event A happening GIVEN that B has already happened.

    3. probability of a complement
    P(~A)=1-P(A)

    4. identity:
    P(A∨B)=P(A)+P(B)-P(A∧B)

    5. De Morgan's law:
    ~A ∧ ~B \equiv; ~(A ∨ B)
    so
    P(~A ∧ ~B)=1-P(A ∨ B)



    We are given
    P(A)=1/600;
    => P(~A)=1-1/600=599/600;

    P(B|A)=0.9;
    => P(B∧A)/P(A)=0.9
    => P(B∧A)=0.9/600=11500=3/2000
    P(B|~A)=0.1;

    => P(B∧~A)/P(~A)=0.1
    => P(B∧~A)=0.1*P(~A)=599/6000

    This also means that
    P(B∧A)+P(B∧~A)=3/2000+599/6000
    =P(B∧(A∨~A)=608/6000
    =P(B)
    =608/6000
    =38/375
    => P(~B)=1-P(B)=1-38/375=337/375
    (a) find P(A|B)
    P(A|B)
    =P(A∧B)/P(B)
    =P(B∧A)/P(B)
    Substitute P(B∧A) and P(B)
    to solve for P(A|B)

    Hint: answer to part (a) is much greater than P(A).

    (b) find P(~A|~B)
    P(~A|~B)
    =P(~A∧~B)/P(~B)
    =(1-P(A∨B))/P(~B)
    =(1-(P(A)+P(B)-P(A∧B))/P(~B)

    All the quantities P(A), P(B), P(A∧B) and P(~B) have been calculated previously, so just substitute to find the answer.

    Hint: answer for part B is very close to 1.

    Finally, if you need detailed explanations or discussions, please post.

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