Question regarding Bayes rule I know that Bayes rule states
$P(X|Y) = P(X  and  Y)/ P(Y)$
What happens if P(X|X>=Y)?
Will I divide by marginal probability of Y or P(X>=Y). For instance how would I approach problem given following PMF
X/Y          1            4
1            0.018        0.153
2            0.045        0.18
5            0.235        0.369

 A: In the formula 
$$ P(A | B) = \frac{P(A \cap B)}{P(B)}$$
$A$ and $B$ are, to start with, events (of which $B$ must have non-zero probability).  By a slight abuse of notation, we write the same when $X,Y$ are random variables.
$$ P(Y | X) = \frac{P(X\cap Y)}{P(X)}$$
but here the events are to be understood as the particular values that each variable can take . That is, by $P(X)$ we mean $P(X=x)$, and so on. This can also be generalized to continuous variables, using densities. All this is explained in the Wikipedia.
The expression $P(X|X \ge Y)$ has no problem, you just need to remember that the condition is an event. If you feel unsure, write it more explicitly:
$$P(X=x|X \ge Y)=\frac{P(X=x \cap X \ge Y)}{P(X \ge Y)}$$
From your table, the numerator and denominator should be easily computable, by summing the corresponding terms.
This should also make you clear that $P(X|Y)$ depends on two variables (in your example, you should compute $6$ values, while $P(X|X \ge Y)$ depends only on $X$ (you must compute $3$ values)
