**Executive summary:** If you look at the whole Hilbert scheme associated to a given polynomial, the locus of points corresponding to nonsingular (which I take to mean smooth) subschemes can sometimes be very small in terms of dimension and number of irreducible components. So in this sense, most subschemes are singular.

**Details:** The Hilbert scheme $\operatorname{Hilb}^P_{\mathbf{P}^n}$ associated to a given Hilbert polynomial $P$ is connected (a theorem of Hartshorne), but in general it has many irreducible components, each with its own generic point. Thus there are several different "generic" closed subschemes with the same Hilbert polynomial, each a member of a different family.

The locus of points in the Hilbert scheme corresponding to *smooth* (=nonsingular) subschemes of $\mathbf{P}^n$ is a Zariski open subset, which implies that it is Zariski dense in the union of the components that it meets, but there are often other components of the Hilbert scheme all of whose points correspond to singular subschemes.

Because the Hilbert scheme need not have a single generic point, one might ask: How many of these generic points parametrize singular subschemes, and what are the dimensions of the corresponding components of the Hilbert scheme?

As a case study, consider the Hilbert scheme $H_{d,n}$ of $d$ points in $\mathbf{P}^n$, i.e., the case where $P$ is the constant polynomial $d$. Points of $H_{d,n}$ over a field $k$ correspond to $0$-dimensional subschemes $X \subseteq \mathbf{P}^n$ of length $d$, or in other words, such that $\dim_k \Gamma(X,\mathcal{O}_X) = d$. Each smooth $X$ with this Hilbert polynomial is a disjoint union of $d$ distinct points. These smooth $X$'s correspond to points of an irreducible subscheme of $H_{d,n}$, and the closure of this irreducible subscheme is a $dn$-dimensional irreducible component $R_{d,n}$ of $H_{d,n}$. Sometimes $H_{d,n}=R_{d,n}$, which means that every $X$ is smoothable. But for each fixed $n \ge 3$, Iarrobino observed that $\dim H_{d,n}$ grows much faster than $\dim R_{d,n}$ as $d \to \infty$. (He proved this by writing down large families of $0$-dimensional subschemes, like $\operatorname{Spec} (k[x_1,\ldots,x_n]/\mathfrak{m}^r)/V$, where $\mathfrak{m}=(x_1,\ldots,x_n)$ and $V$ ranges over subspaces of a fixed dimension in $\mathfrak{m}^{r-1}/\mathfrak{m}^r$.) This shows that $H_{d,n}$ is not irreducible for such $d$ and $n$, and that the ``bad'' components all of whose points parametrize singular subschemes can have much larger dimension than the one component in which a dense open subset of points parametrize smooth subschemes. With a little more work, one can show that the number of irreducible components of $H_{d,n}$ can be arbitrarily large (and as already remarked, the components themselves can have larger dimension than $R_{n,d}$). So in this sense, one could say that for $n \ge 3$, most $0$-dimensional subschemes in $\mathbf{P}^n$ are singular.

For more details about $H_{d,n}$, including explicit examples of nonsmoothable $0$-dimensional schemes, see the following articles and the references cited therein:

The moduli space of commutative algebras of finite rank

Hilbert schemes of 8 points

(Warning: my notation $H_{d,n}$ is different from the notation of those articles.)