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.



Is there a precise notion of "almost all" such that almost all finite groups are Galois groups of extensions of the rationals?
Joel's comment above, citing the result of Shafarevich, means that this question can be answered if one can prove that



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> almost all finite groups are solvable.
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for some sense of "almost all". For this, I refer you to to this paper:



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> 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. 
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The MathSciNet review, by Koichiro Harada, explains the result:



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> 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 $\tau(G)=|G|^{−1}\prod|H/K|$, 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.
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