Probability of $x$ being rational Let the sequence $(x_1, x_2,...)$ be generated by sampling (uniformly and independently) the integers $\{0,1,2,...,9\}$.
From the sequence we construct the real number $x =  0.\overline{x_1x_2x_3..}$
What is the probability that $x$ is rational?
I just came up with this question, and realized that I do not even know how to start thinking about it. Intuitively, since the rationals are countable and the irrationals are uncountable, one might guess that the probability of $x$ being rational is 0.
Does the question even make sense from a probability theory's perspective?

The set of all rational numbers in $[0,1]$ is countably infinite.  So by Axiom 3 of probability theory, we can just add up the probabilities one by one.  Let $X$ be your random variable.  Then
$$ P[X \in [0,1]\cap \mathbb{Q}] = \sum_{x \in [0,1]\cap \mathbb{Q}} \underbrace{P[X=x]}_{0} = 0 $$
where we use the fact that $P[X=x]=0$ for all $x \in [0,1]$ (indeed, a comment by Ivan mentions that $X$ is uniformly distributed over $[0,1]$). The same is true for any random variable $X$ with a continuous CDF:  The probability that $X$ is irrational is 1.
For your situation, you can see that $P[X=x]=0$ for all $x \in [0,1]$ by noting that for your i.i.d. random variables $\{X_i\}_{i=1}^{\infty}$ we have $X= \sum_{i=1}^{\infty} X_i 10^{-i}$.
Now a prior comment notes that every $x \in [0,1]$ has at most two representations $0.a_1a_2a_3...$ (there can be two if one has an infinite tail of 9s).  But we observe that for any particular sequence $\{a_1, a_2, a_3, ...\}$ such that $a_i \in \{0, 1, 2, ..., 9\}$ for all $i$ we get
\begin{align}
P[X_1=a_1, X_2=a_2, X_3=a_3, ...] &= P[\cap_{i=1}^{\infty} \{X_i=a_i\}] \\
&\overset{(a)}{=} \prod_{i=1}^{\infty} \underbrace{P[X_i=a_i]}_{1/10} \\
&= 0
\end{align}
where step (a) holds by mutual independence of $\{X_i\}_{i=1}^{\infty}$.
