If the cardinality of two infinite sets is different are also its elements' probabilities different when selecting from its union? [closed]

Suppose we have two infinite sets $A$ and $B$ such that $|A|>|B|$. And let $C=A\cup B$.

Assume that $P(x)$ is the probability of selecting an element $x, \forall x \in C$ following the uniform distribution.

Is it true that:

$$P(x_a\mid x_a \in A) \;>\; P(x_b \mid x_b \in B)$$

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Define the uniform distribution on $C$ and $P(x_a|x_a\in A)$, just to start with. By the way, if $A$ and $B$ are finite the uniform distribution over $C$ is well-defined, and $P(A)>P(B)$ whenever $|A|>|B|$. I do not, though, if it's relevant w.r.t. your question. – Ilya Aug 21 '12 at 13:18
Hi @Ilya, thanks for your comment. Yes the finite case is relevant. However I have no idea how to define a uniform distribution over C, I mean all elements of C has the same probability to be selected? – Jose Antonio Martin H Aug 21 '12 at 13:29
Well, if $C = \Bbb N$ what is the probability to select $x = 42$? – Ilya Aug 21 '12 at 13:33
Yes, I see the point, is it completely necessary to define this to answer P(A)>P(B)?. – Jose Antonio Martin H Aug 21 '12 at 13:37
I'm afraid, yes - uniform distributions are defined only over very special sets like: finite sets, and bounded intervals $[a,b]$. What I can say is that very often (though, not always) if for two infinite sets it holds that $|A|>|B|$, and the probability measure (no necessarily uniform) is defined over $C = A\cup B$ then $P(A) = 1$ and $P(B) = 0$. But in general, of course, it all depends on the way you define the measure. As an example, quite often I have to deal with a probability measure $P$ over $\Bbb R$ such that $P(\{0\}) = P(\Bbb R\setminus \{0\}) = \frac12$. – Ilya Aug 21 '12 at 13:44
For shure if $A\subseteq B$ then P($X_{a}$ ∣$X_{a}$ ∈A)>P($X_{b}$ ∣$X_{b}$ ∈B)
 Hi @Hernan, but who is saying that $A \subseteq B$? – Jose Antonio Martin H Aug 21 '12 at 13:31 I am only wrote that it is true if $A\subseteq B$ in the general case I am not shure. – Hernan Aug 21 '12 at 13:34 Are you sure?, don't you think you should write $B\subseteq A$ instead ? – Jose Antonio Martin H Aug 21 '12 at 13:39 Or even $B \subset A$ ? – Jose Antonio Martin H Aug 21 '12 at 13:40 yes you are right my mistake – Hernan Aug 21 '12 at 14:43