I vainly tried to define a notion of "almost solvable group" such that every "almost solvable group" is the Galois group of a finite extension of the rationals, but I can't figure out the right way to do so, so I want to ask a probably less ambitious question. Is there a precise notion of "almost all" such that almost all finite groups are Galois groups of finite extensions of the rationals? If so, what is the considered notion exactly?
Thanks in advance.

10$\begingroup$ Since "almost all finite groups are 2groups" in the sense that the proportion of isom. classes of groups of order up to $x$ that are $2$groups goes to $1$ when $x$ goes to infinity, and since all $2$groups are nilpotent, hence solvable, and since all solvable have been shown by Shafarevich to be Galois groups of finite extensions over $\mathbb Q$, then the answer to your question is yes. $\endgroup$– JoëlApr 28, 2014 at 19:45

4$\begingroup$ @Joël Is "almost all finite groups are $2$groups" actually known? I was under the impression it was still open. $\endgroup$– Alex BeckerApr 28, 2014 at 19:56

$\begingroup$ Alex, you're right, I checked on wikipedia and it is said to be "a folklore conjecture". So my comment is wrong. I leave it as it may help precise the question: it proposes a natural notion of "almost all finite groups" for which the Galoistheoretic question "are almost all finite groups galois groups over $\mathbb Q$?" can be reduced to a Grouptheoretic conjecture. Is it possible to prove with current technology that almost all finite groups are Galois groups over $\mathbb Q$ in that sense ? $\endgroup$– JoëlApr 28, 2014 at 20:04

1$\begingroup$ Proving almost all finite groups are 2groups looks hard, but perhaps it is reasonable to assert that almost all finite groups are soluble, since nonabelian simple groups are extremely sparse in the class of finite groups and don't admit many extensions. $\endgroup$– Colin ReidApr 29, 2014 at 3:34

$\begingroup$ Derek Holt makes some interesting comments about conjectured properties of "almost all finite groups", in his answer to this question: mathoverflow.net/questions/164202/… you might find it of interest... $\endgroup$– Nick GillApr 29, 2014 at 15:33
1 Answer
Joel's comment above, citing the result of Shafarevich, means that this question can be answered if one can prove that
almost all finite groups are solvable.
for some sense of "almost all". For this, I refer you to to this paper:
Camina, A. R.; Everest, G. R.; Gagen, T. M., Enumerating nonsoluble groups—a conjecture of John G. Thompson. Bull. London Math. Soc. 18 (1986), no. 3, 265–268.
The MathSciNet review, by Koichiro Harada, explains the result:
It seems obvious that there are more solvable groups than nonsolvable ones. J. Thompson has made a conjecture, expressing the rarity of nonsolvable groups in a precise way. For a given $G$, let $\tau(G)$ be the number defined as $\tau(G)=G^{−1}\prodH/K$, where $H/K$ ranges over all nonabelian composition factors of $G$. Obviously $0\leq\tau(G)\leq 1$ and $\tau(G)=0$ if and only $G$ is solvable. For a given $\varepsilon>0$, let $f_\varepsilon(n)$ denote the number of isomorphism classes of groups $G$ with $G\leq n$ and $\tau(G)\geq\varepsilon$. Then $f_0(n)$ is the number of groups $G$ with $G\leq n$. Thompson's conjecture is $f_\varepsilon(n)=o(f_0(n))$ as $n\to\infty$ for all $\varepsilon>0$. The conjecture is proved in this paper.

$\begingroup$ For a finite $p$group $P$, one can consider the class of all finite groups (up to isomorphism) containing $P$ as a Sylow $p$subgroup; let us call this the orbit of $P$. It is shown by Henn and Priddy that almost all finite groups are $p$nilpotent in the sense " for almost all $p$groups $P$ of Frattini class $n$, any group in the orbit of $P$ is $p$nilpotent". More precisely, If one denotes by $A_d$ the set of $p$groups of Frattini class $n$that can be generated by $d$ elements, and one denotes by $B_d$ the subclass of $A_d$ having the property that any member $P$ of $B_d$, $\endgroup$ Apr 30, 2014 at 21:52

$\begingroup$ has an orbit in which all the members are $p$nilpotent, then $B_d/A_d \rightarrow 1$ when $d \rightarrow \infty$. It is interesting to extend this definition to include all the primes. $\endgroup$ Apr 30, 2014 at 21:58