What is the probability of randomly chosen real number is even? I read in the book A First Course in Probability by Sheldon Ross the following statement:

Technical Remark. We have supposed that $P(E)$ is defined for all the events $E$ of the sample space. Actually, when the sample space is an uncountably infinite set, $P(E)$ is defined only for a class of events called measurable. However, this restriction need not concern us as all events of any practical interest are measurable.

The set of Real numbers is an infinitely uncountable set then how can we calculate the probability of even real number?

This gets into a branch of math called measure theory. The idea is, given a set of real numbers, can we give certain measure to it? Probably the most common measure (as far as I know), is Lebesgue measure. It extends the idea of lengths of intervals. For example, any interval $[a,b]$, has measure $b-a$. This is simple enough, but more complicated sets are more complicated to measure. Some sets can not be measured. In a Probability space $S$,  we can then say the probability of an event $E$ is $\frac{\mu(E)}{\mu(S}$. Here $\mu$ is the measure function. It turns out in this case $\mu(E) = 0$ and $\mu(S) = \infty$, so the probability is 0.
If you want to see better why this works, the basic requirements of a measure are


*

*Non-negativity: For all E $\mu(E) \ge 0$.

*Null Empty Set: $\mu(\emptyset) =0$.

*Countable Additivity: For a countable collection of pairwise disjoint sets $\{E_i\}$, $\mu(\bigcup\limits_{i=1}^{\infty} E_{i}) = \sum_{i=1}^{\infty} \mu(E_i)$.


The measure of any point is $0$, as as the integers are a countable collection of points, their measure is the sum of a bunch of zeros, and so is zero.
We can write the reals as an infinite union of intervals of length one, and thus the reals have measure $\infty$.
If you want a more detailed explanation of measure the wikepedia page does a pretty good job explaining it, and it is short. https://en.wikipedia.org/wiki/Measure_(mathematics).
