Is there a set of primes such that the union of their multiples occurs with frequency $1/n$, where $n$ is an integer? 
Are there two or more primes, $p_1, p_2, \ldots$ such that the union of their multiples occurs with frequency $1/n$, where $n$ is an integer? If so, can that n be prime? 

I put it into a computer and pairs of primes seem to get arbitrarily close (looked at primes < 10000), but I didn't see any. 
(Please don't hurt me, I know I'm bad at math.)
 A: I am interpreting your question to be asking about the (asymptotic) natural density of the set
$$
\{ p_1, 2p_1, 3p_1, \dots \} \cup \{ p_2, 2p_2, 3p_2, \dots \} \cup \cdots \cup \{ p_k, 2p_k, 3p_k, \dots \},
$$
where $p_1<\cdots<p_k$ are distinct primes.
This union is a periodic set with period $p_1\cdots p_k$, and so its natural density is equal to its number of elemnets up to $p_1\cdots p_k$ divided by $p_1\cdots p_k$.
It's not hard to show by induction on $k$, using the fact that $p_1,\dots,p_j$ each divide $n$ if and only if $p_1\cdots p_k$ divides $n$, that the complement of this set has natural density
$$
\prod_{j=1}^k \bigg( 1-\frac1{p_j} \bigg);
$$
thus the density of your set is $1$ minus this quantity. In particular, this density is a fraction whose denominator is divisible by the largest prime $p_k$.
Therefore we can answer "no" to your second question: if this density is to be the the reciprocal of a prime, then it has to equal $1/p_k$. But note that the equation
$$
1 - \prod_{j=1}^k \bigg( 1-\frac1{p_j} \bigg) = \frac1{p_k}
$$
is algebraically equivalent to
$$
\prod_{j=1}^{k-1} \bigg( 1-\frac1{p_j} \bigg) = 1;
$$
since each factor is less than $1$, this is impossible when $k\ge2$.
