The inverse Galois problem is a classical problem in mathematics and asks whether every finite group can be realized as the Galois group of a finite field extension of the rational numbers. The analogous question has been answered for function fields of a single variable over an algebraically closed field of characteristic zero. In the number field case, Shafarevich showed that any finite solvable group can be realized as the Galois group of a field extension over $\mathbb{Q}$.

In particular, we note that every $p$-group is solvable. More specifically, any group whose order is $2^k$ for some $k \in \mathbb{N}$ is solvable. Thus, any group of order $2^k$ is the Galois group of some number field over $\mathbb{Q}$. In a conversation with Bjorn Poonen in July, he mentioned that if there was a 'finite group' race, where he would list all the groups (up to isomorphism) of order $2^k \leq x$ and another person listed every other group whose order is at most $x$, then he would surely list more groups; and in fact, he said that his list will have a proportion approaching 100% as $x \rightarrow \infty$.

Applying Poonen's heuristic to the inverse Galois problem, then order groups by their order, a proportion approaching 100% of all finite groups are Galois groups of number fields over the rationals.

Has this been proved? Here, I am satisfied with a much weaker notion of 'most' which is the statement that a proportion exceeding $1/2$ of all groups of order up to $x$ are Galois groups as $x \rightarrow \infty$.

Edit: To clarify, I am asking whether statement 'a proportion exceeding $1/2$ of all finite groups are Galois groups over $\mathbb{Q}$' has been proved or not. This statement is implied by the statement `most finite groups have order $2^k$ for some $k$' by Shafarevich's theorem.