Count the number of prime factors of a number $n$ to include multiplicity, so that $$n=24=2^3 \cdot 3 = 2 \cdot 2 \cdot 2 \cdot 3$$ has $4$ prime factors, and $$n = 6500 = 2^2 \cdot 5^3 \cdot 13 = 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5 \cdot 13 $$ has $6$ prime factors.
The distribution is quite regular. Here it is for $n \le n_\max$, $n_\max=10^7$:
[![DistPrimeFacts][1]][1]
About a quarter of $n \le 10^7$ have $3$ prime factors.
Q. What is this distribution explicitly? Where is its peak, for $n \le n_\max$?