Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Decompostion of conditional probabilities

  1. Mar 24, 2012 #1
    So If B=B1UB2U.....UBn would Pr(A|B)=Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B2)+.....+Pr(A|Bn)Pr(Bn)?

    I haven't found a formula for this but it makes intuitive sense.
  2. jcsd
  3. Mar 24, 2012 #2


    User Avatar
    Science Advisor

    Hey torquerotates.

    Before I give my take on the answer, do you assume that all the B1, B2, etc are disjoint? In other words is Bi AND Bj = NullSet for i != j?
  4. Mar 24, 2012 #3
    Yes they are disjoint.
  5. Mar 24, 2012 #4


    User Avatar
    Science Advisor

    For two events I get the following:

    P(A|B1UB2) = P(A and (B1UB2))/P(B1UB2) = P((A and B1) U (A and B2))/P(B1UB2) = P(A and B1)/P(B1UB2) + P(A and B2)/P(B1UB2) - P(A and B1 and B2)/P(B1UB2)

    = P(A and B1)/(P(B1) + P(B2)) + P(A and B1)/(P(B1) + P(B2)) - 0

    = 1/P(B)[P(A and B1) + P(A and B2)]

    = 1/P(B)[P(A and B1]P(B1)/P(B1) + P(A and B2)P(B2)/P(B2)]

    = 1/P(B)[P(A|B1)P(B1) + P(A|B2)P(B2)] if we assume P(Bi) > 0 and is a valid probability.

    If the above is right then your answer is wrong in general but right when B is the universal probability space.

    Because your probability P(B) may not be one, it means that the above needs to be corrected using this information. Just for an example consider P(B1) = P(B2) = 1/4 and P(A|B1) = P(A|B2). But since 1/4[P(A|B1) + P(A|B2)] = 1/2P(A|B) for this example since P(B1) = P(B2) and since B1 and B2 are disjoint we have:

    Then under these conditions we have:

    1/P(B) x [P(A|B1)P(B1) + P(A|B2)P(B2)]
    = 1/(1/2) x [P(A|B1)x1/4 + P(A|B1)x1/4]
    =(1/4)/(1/2) x [P(A|B1) + P(A|B2)]
    = 1/2 x [(1/2)/(1/4)]xP(A|B)
    = 1/2 x 2 x P(A|B)
    = P(A|B) for this example.

    If we did not 'normalize' by P(B) we would not have gotten the right answer.

    While I think the above is right, I would like others to check if they could just to make sure, but I'm only using a few identities which are basically distribution over sets and multiplication by 'x/x' terms for P(Bi)/P(Bi).
  6. Mar 25, 2012 #5

    Stephen Tashi

    User Avatar
    Science Advisor

    Did you do any examples?

    A = the roll of a fair die produces an even number
    B = the face that comes up is a 1 or a 2
    B1 = the face that comes up is a 1
    B2 = the face that comes up is a 2
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook