If $\pi_k(n)$ is the cardinality of numbers with k prime factors (repetitions included) less than or equal n, the generalized Prime Number Theorem (GPNT) is:
$\pi_k(n)\sim \frac{n}{\ln n} \frac{(\ln \ln n)^{k-1}}{(k-1)!}.$
The qualitative appearance of the actual distribution of $\pi_k(n)$ for k = 1,2,3,..., agrees very well with the GPNT, for numbers $n$ within reach of my laptop. But I noticed that as $n$ and $k$ get large, "most" of the numbers less than $n$ seem to have relatively few factors.
Writing $n = 2^m$ and replacing $k$ by $x$ we can graph
$f(x) =\frac{2^m (\ln\ln 2^m)^{x-1}}{\ln 2^m (x-1)!}$
from $x = 1$ to $m$ (since no number will have more than m factors) and see that for relatively small fixed $m$, most of the area under the curve f is is contained in a steep bell-shaped curve on the far left of the image.
I take this to suggest that as we consider very large sets, $S_m = \{ 1,2,3,...,2^m\},$ almost all elements of these sets have a "very small" number of factors (including repetitions).
Can this idea be (or has it been) quantified? The phrase "very small" is frustrating, and I think we might be able to say something more concrete about, for example, concentration of the proportion of area as a function of x and m...? Thanks for any suggestions.
Edit: the answer Eric Naslund gave below is splendid and I won't neglect to accept it. In response to the answer, I wonder if there is any reason not to be able to get something like that answer from the expression $f(x)$?
After all, $f(x)$ appears to be a Poisson-like curve with a mean near the average number of prime factors. If I let m = 100 and then 500 (i.e., we're using $2^{100},2^{500}$), $f'(x) = 0$ at $x \approx 4.73, 6.34$, respectively, while $\ln\ln 2^m$ is respectively 4.23, 5.84. If f is a valid expression for the asymptotic behavior of $\pi_k(n)$, wouldn't we expect it to give us this additional information? Can we not prove it?