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Homework Help: Independence and conditional probability

  1. Jul 20, 2006 #1
    if X and Y are events which are independent of each other, but neither are independent with A,

    is this equality true for conditional probabilities:
    P( X, Y | A) = P(X|A) * P(Y|A)

    if not,
    how do you solve for P(A | X,Y)
    given that you only know P (A) and P(X|A) and P(Y|A)?

    The reason I came up with the above probability where I have:
    [tex] P(A| X, Y) = \frac {P(X, Y | A) P(A)}{P(X, Y |A) P(A) + P(X, Y | A^c) P (A^c)} [/tex]
    is that I used Baye's Thm.

    Note: P(X, Y |A) is not given.
  2. jcsd
  3. Jul 20, 2006 #2


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    Science Advisor

    First, by P(X,Y) do you mean P(X and Y) or do you mean P(X or Y)?

    Assuming that you mean P(X and Y) then, yes, since X and Y are independent, P(X)= P(Y) so P(X and Y| A)= P(X|A)P(Y|A).
  4. Jul 20, 2006 #3
    Cheers for the help, HallsofIvy.

    well, this is how it is stated in the problem:
    X is the event of testing negative in a drug test
    Y is the event of testing positive in the drug test
    A is the event of having a disease.

    Given that a person went to have a drug test three times, testing positive once and negative twice. What is the probability he has a disease?

    here, X and Y are mutually exclusive events. However, are [tex]X_1[/tex] and [tex]X_2[/tex] mutually exclusive? It would seem that testing thrice would mean the question is asking for the union of the 3 events. Is this correct?

    so would the solution be
    [tex] P(A| X, X, Y) = \frac {P(X|A)P(X|A)P(Y|A)P(A)}{P(X|A^c)P(X|A^c)P(Y|A^c)P(A^c)}[/tex] ?
    Last edited: Jul 20, 2006
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