Wait, would not "any even number" have the same probability to be picked then? Equals zero. That does not seem right. Is there something "wrong" with the question, that makes it different from "what proportion of all integers have the property X"?
If you were to pick a random integer between 0 and infinity (which is indeed not an integer, it's not even a number), the chance of it being even is 0.5
+. 0 is an even number so in any set of consecutive integers starting with 0, there is either equal number of even and odd numbers, or one more even than odd.
With primes, it gets more interesting. We could try it the brute-force way. We know that in an infinite number of positive integers, there is an infinite number of numbers. Also, an infinite number of those are primes.
This results in the chance of getting a prime being infinity/infinity, which is undefined.
So let's try a different way.
The Prime Number Theorem says that the number of primes equal to lower than R (and we can easily substitute N for R) is roughly R/(ln (R)).
Since we are operating in natural numbers, that becomes N/(ln(N)). The number of natural numbers equal to or lower than N is N.
So the chance of a random number between 1 and N being a prime is (N/(ln(N))/N), which can be simplified to 1/(ln N).
Since ln (N) limits to infinity at infinity, 1/ln(N) limits to 0 at infinity.
As such, the chance of a random integer between 1 and infinity is indeed 0
+, a chance so low, that it might as well be 0, but not actually 0.