If a compound is more solube than another compound, can we say that it is more likely to stay in a given solution than the other less soluble compound would be to stay in that same given solution?

I don't quite understand your question... I will however tell you something, that I recently learned in my classes (you lucky! ;))

Ok, every substance has a constant of solution, a sort of rule for it's extent of dissolution in water and other solvents.

Let's say a certain compound AB.

A^+ B^- (s) \rightleftharpoons A^+ (aq) + B^- (aq) (adding water)

The constant is called K_sp.

It is given by:

K_{sp} = \frac{[A^+][B^-]}{[AB]}

The square brackets mean concentration, and since a solid cannot have a concentration, we make it 1. Hence,

K_{sp} =[A^+][B^-]

Ok, you'll notice the reversible sign up there, in the equation. It's because the solid is in dynamic equilibrium with the ions, hence explaining why some compounds are more soluble than others. They have different equilibrium points.

Back to the subject, the more soluble a compound is, the more ions you'll have and the higher the constant you'll have, and thus that constant is closely related to the solubility.

Therefore, any compound will obey that constant at a given temperature and this will determine how many ions will stay in the solution or not. A more soluble compound will have more ions in the solution than a less soluble one, given the ions are in equilibrium with the solid. If there is no solid, some of the ions will become solid to keep the equilibrium.