Is probability-raising closed under union? Suppose that $\Pr(X \mid A) > \Pr(X)$, and that $\Pr(X \mid B) > \Pr(X)$.  Does it follow that $\Pr(X \mid A \cup B) > \Pr(X)$?
$\Pr(X \mid A \cup B) > \Pr(X)$ holds just in case 
$$
[\Pr(X A) - \Pr(X) \cdot \Pr(A)] + [\Pr(XB) - \Pr(X) \cdot \Pr(B)] > \Pr(X A B) - \Pr(X) \cdot \Pr(A B)
$$
($XA$ is the intersection of $X$ and $A$).  Both of the differences on the left-hand-side are positive, so the left-hand-side is positive.  But the difference on the right-hand-side could also be positive, and I don't see why it couldn't be more positive than the sum on the left.  I went looking for simple counterexamples, but couldn't find any. 

Here is a counterexample. Two balls are drawn, with replacement, from an urn containing
$1$ black ball and $2$ white balls. Consider the following events.
$X$: The two balls are the same color.
$A$: The first ball is white.
$B$: The second ball is white.
Then $\Pr(X)=\frac59$, $\ \Pr(X\mid A)=\Pr(X\mid B)=\frac23\gt\Pr(X),$ and $\Pr(X\mid A\cup B)=\frac12\lt\Pr(X).$
Intuitively, without doing the calculations: Drawing a white ball on (say) the first draw improves the chances of matching colors, since there are more white balls than black. However, the event "at least one white ball" worsens the chances of a match, since all we are eliminating is a favorable case, two black balls.
