If I were to look at the decimal representation of some irrational or even transdental number,
and if I choose a natural number at random can I expect that it is some digit with probability $0.1$ ?
In general all digits of an irrational number do not have to occur with equal probability. A quick example would be the number $$0.1011011101111011111...$$
Here, the number is composed of only $1$'s and $0$'s, so the probability for digits 2-9 is zero. In fact, depending on how you define the probability of finding a digit, the probability of randomly selecting $0$ from the digits might also be zero. This is the case if you define the probability for the infinite decimal representation to be the limit as $n\rightarrow\infty$ of the probability of finding the digit in a truncated approximation of the number with $n$ digits.
Even with this definition, you could get irrational numbers where this limit doesn't exist for some digits. An example would be something like $$0.101100111111000000...$$ where after each run of $0$'s you keep appending a $1$ until the probability of finding zero in the truncated representation is less than $1/4$, and then you go back to appending $0$'s until the probability of finding a $0$ has gone back up to $1/2$. In this case, our definition for the probability of finding $1$ or $0$ in the infinite decimal representation doesn't give an answer, because the limit doesn't exist.
So I guess the point is that probabilities are a bit tricky to define for infinite sets of numbers, but in general irrational numbers do not have to have equal probabilities for each digit.
Presumably by "at random", you mean "uniformly at random". There is no uniform distribution on the natural numbers; see Probability of picking a random natural number.