torquerotates said:
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).