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

closed as not a real question by Did, William, tomasz, Martin Sleziak, J. M.Oct 2 '12 at 14:15

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

For shure if $A\subseteq B$ then P($X_{a}$ ∣$X_{a}$ ∈A)>P($X_{b}$ ∣$X_{b}$ ∈B)

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 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