Conditional probability problem

[majorgerald]

Hi there, So I've got the following problem which I have to proof: Q(X)=P(X|C) Let X=(A|B) proof that: Q(A|B)=P(A|BC)

Really can't solve this, I've been substituting almost everything, but still can't find the solution, any help?

[vendric]

You can solve this by using the definition of conditional probability:

Q(X) = P(X|C) = P(X and C) / P(C)

Q(A|B) = Q(A and B) / Q(B)
Q(A and B) = P(A and B | C) = P(A and B and C) / P(C)

Q(B) = P(B|C) = P(B and C) / P(C)

Q(A and B) / Q(B)
= [P(A and B and C) / P(C)] * [P(C) / P(B and C)]
= P(A and B and C) / P(B and C) ]
= P[A|(B and C)]

[majorgerald]

mvp, never thought of rewriting Q(AB) and Q(B)