primes and semiprimes found with $n=11,13,17,19 \mod 30$. For $n=11,13,17,19 \mod 30$ what percentage of all primes and all semiprimes less than $100, 1000, 10000...$ will be produced by $n$?
Up to $20,000$ I found $50\%$ were primes and more than $50\%$ were semiprimes. If one continued the number of trials, would the results be closer to primes $10\%$ and semiprimes $20\%$? 
 A: There are two possible readings of your question (ignoring the semiprimes bit):


*

*what proportion of the primes up to $n$ are congruent to 11, 13, 17, or 19 mod 30?

*what proportion of the positive integers up to $n$ congruent to 11, 13, 17, or 19 mod 30 are primes?


The prime number theorem for arithmetic progressions answers both of these.  The total number of primes up to $n$ is approximately $Li(n)$, where $Li(n)$ is the so-called logarithmic integral,
$$Li(n) = \int_2^n {dt \over \log t}.$$
The total number of primes up to $n$ congruent to $a$ mod 30, where $a$ is relatively prime to 30, is approximately
$$ {1 \over \phi(30)} Li(n) $$
where $\phi$ is the Euler totient function -- that is, $\phi(n)$ is the number of positive integers less than $n$ and relatively prime to $n$.  In particular $\phi(30) = 8$, the size of the set $\{ 1, 7, 11, 13, 17, 19, 23, 29 \}$.  
So one-eighth of all primes up to $n$ are congruent to 11 mod 30, and similarly for 13, 17, or 19.  One-half of all primes up to $n$ are congruent mod 30 to one of these four.
On the other hand, there are approximately $4n/30$ integers less than $n$ which are congruent to 11, 13, 17, or 19 mod 30.  Of these approximately $Li(n)/2$ are prime, so the proportion of such integers which are prime is approximately
$$ {Li(n)/2 \over 4n/30} = {15 \over 4} {Li(n) \over n}$$
in particular $Li(n)$ is approximately $n/(\log n)$ and so this is approximately 
$$ {15 \over 4 \log n}. $$
As $n \to \infty$ this gets as small as you like.  In particular it does not approach a positive constant like your conjectured $10\%$. The proportion of positive integers less than $n$ which are prime is about $1/(\log n)$, so your set is "richer" in primes than the positive integers by a factor of $15/4$.
