Is almost any group generated by two generators? What is the asymptotic probability that a randomly chosen finite group can be presented with $2$ generators? More precisely, what is $$ \lim _{n \to \infty} \frac{\text{number of 2-generated groups of order} \le n}{ \text{number of groups of order} \le n}$$ if it exists?
 A: In fact, almost all groups aren’t. Moreover, a stronger statement is true:

Suppose $d$ is an arbitrary natural number. Then $\lim _{n \to \infty} \frac{\text{number of }d \text{-generated groups of order} \le n}{ \text{number of groups of order} \le n} = 0$

To prove this we need two facts. 
The first one is the Theorem 1 from “Enumerating Boundedly Generated Finite Groups” by Alexander Lubotzky. It states:

There exists a constant $c$, such that for any natural $d$ and $n$ the number of $d$-generated groups of order less than $n$ does not exceed $n^{cd\ln(n)}$

The second one is a theorem proved by Charles Sims in “Enumerating $p$-groups”, that states:

For any prime $p$ and natural $n$ the number of groups of order $p^n$ is $p^{\frac{2}{27}n^3 + O(n^{\frac{8}{3}})}$

Or, to be more exact, its corollary:

There exists such a constant $b$, that for any natural $n$, the number of groups of order strictly less than $n$ is greater than $n^{b(\ln(n))^2}$

From those statements we get:
$$0 \leq \lim _{n \to \infty} \frac{\text{number of }d \text{-generated groups of order} \le n}{ \text{number of groups of order} \le n} \leq \lim_{n \to \infty} \frac{n^{cd\ln(n)}}{n^{b(\ln(n))^2}} = 0$$
