Let $q(x_1,x_2):=ax_1^2+bx_1x_2+cx_2^2$ be a positive-definite quadratic form with $a,b,c\in\Bbb{Z}$.
Let $r_q(n)$ be the number of natural numbers $k\leq n$ that are represented by $q$, i.e., such that $q(x_1,x_2)=k$ for some $x_1,x_2\in\Bbb{Z}$.
Is it necessarily true that the probability of a random number being representable by $q$ is zero, in the sense that $\lim_{n\to\infty}\frac{r_q(n)}{n}=0$?
In the nice example $q(x_1,x_2)=x_1^2+x_2^2$, we have that $r_q(n)$ is proportional to $\frac{n}{\sqrt{\log n}}$ (according to this article), so $\frac{r_q(n)}{n}$ is proportional to $\frac{1}{\sqrt{\log n}}$, which tends to zero.
Less pleasant examples like $q(x_1,x_2)=3x_1+7x_2^2-x_1x_2$ seem to be much more restrictive on possible representations, so it looks like the conjecture should be true.
Given that quadratic forms are intensely studied since Gauss (at least), I imagine this must be something well-known.
$$a\, q(x,y) = 4 a^2 (2x-(-b+\sqrt{b^2-4ac})y)(2x-(-b-\sqrt{b^2-4ac})y) $$
If $b^2-4ac$ is not a square then $q(x,y)$ will be a norm in $\Bbb{Q}(\sqrt{b^2-4ac})$ where half of primes are non-split so the integer norms have density $0$ (and asymptotic $r \frac{n}{\sqrt{\log n}}$).
If $b^2-4ac$ is a square then $a q(x,y)$ is equivalent
either to $x^2-y^2$ if $b^2-4ac\ne 0$ (which has density $1/2$ but is not positive-definite)
or $(x+y)^2$ if $b^2-4ac=0$ which has density $0$ (and asymptotic $\sqrt{n}$) and is only semi positive-definite.