Rewrite probabilities as boolean algebra

I'm given three binary random variables: X, Y, and Z. I'm also given the following:

P(Z | X)

P(Z | Y)

P(X)

P(Y)

I'm then supposed to determine whether or not it is possible to find P(Z | Y, X). I've tried rewriting the solution in the form of Bayes' Theorem and have gotten nowhere. Given that these are boolean random variables, is it possible to rewrite the system in terms of boolean 开发者_运维技巧algebra? I understand that the conditionals can be mapped to boolean implications (x -> y, or !x + y), but I'm unsure how this would translate in terms of the overall problem I'm trying to solve.

(yes, this is a homework problem, but here I'm much more interested in how to formally solve this problem than what the solution is...I also figured this question would be entirely too simple for MathOverflow)

I'll bet someone has done this more elegantly, but...

In this case, no, it's not possible to determine P(Z|Y,X). In general I think it's possible to start with a set of independent "atomic" probabilities and eliminate them as we add constraints. For example, looking at X and Y, we start with four probabilities:

P( X,  Y) = a
P( X, ~Y) = b
P(~X,  Y) = c
P(~X, ~Y) = d

Now we add the constraint that the probs must add up to 1. We can eliminate one variable, any variable, say d:

P( X,  Y) = a
P( X, ~Y) = b
P(~X,  Y) = c
P(~X, ~Y) = 1-a-b-c

Now suppose we also know P(X)=K:

P( X,  Y) = a
P( X, ~Y) = K-a
P(~X,  Y) = c
P(~X, ~Y) = 1-K-c

And so on. In the stated problem we can eliminate five of the original eight probabilities, but then we're asked for the ratio of two that are still independent.