>>13733843This is the correct answer. But we can still answer the question in a weak sense with some assumptions.
I'm going to assume by integer, you really mean nonnegative integer. In that case, we can consider the set of integers from 0 to n, inclusive. There are (n+1)^2 ways to make a selection of two elements from this set. Furthermore, note that the possible sums which can be made range from 0 to 2n, and the frequency of combinations giving each of these sums forms a pyramid. Then for all prime p less than or equal to n, there are precisely p + 1 ways to sum to them. For all primes q greater than n but less than or equal to 2n, there are 2n - q + 1 ways to sum to them. So for the set of integers from 0 to n, we have a probability of
summing to a prime number. Now, if you take the limit as n goes to infinity, this proportion will go to 0, but in a non-obvious way. It just has to do with the fact that the prime-counting function has logarithmic growth, which affects our numerator's growth, whereas the denominator has steady polynomial growth.
