Random algebraic numbers are linearly disjoint almost surely? It is well-known that if one considers a “random” monic polynomial of fixed
degree, say $X^n+\sum_{k=0}^{n-1}a_kX^k$ where $(a_0,a_1,\ldots, a_n)$ is drawn
from the discrete uniform distribution on $[-N,N]^{n+1}$, then this polynomial
will be irreducible and have Galois group $S_n$ “almost surely”, i.e. the 
probability of this event tends to $1$ when $N\to \infty$.
Now, suppose one considers two random monic polynomials 
$P=X^n+\sum_{k=0}^{n-1}a_kX^k$ and $Q=X^m+\sum_{k=0}^{n-1}b_kX^k$ where
$(a_0,a_1,\ldots, a_n,b_0,\ldots,b_m)$ is drawn
from the discrete uniform distribution on $[-N,N]^{n+m+2}$. Is it also
true that for any root $\alpha$ of $P$ and any root $\beta$ of $\mathbb Q$,
the extensions ${\mathbb Q}(\alpha)$ and ${\mathbb Q}(\beta)$ will be linearly
disjoint over $\mathbb Q$ “almost surely” in the sense above ?
 A: Here is a proof when $m$ and $n$ are distinct.  As you say, “almost surely” the Galois groups of $P$ and $Q$ are $S_n$ and $S_m$, respectively.  Assume without loss that $m<n$.  Let $K$ be the splitting field of $P(X)$ over $\mathbf{Q}$.  Then $K(\beta)/\mathbf{Q}(\beta)$ is Galois, and its Galois group is isomorphic to the Galois group of $K/L$, where $L:=K\cap\mathbf{Q}(\beta)$.  But since
$$
[L:\mathbf{Q}]\le [\mathbf{Q}(\beta):\mathbf{Q}] = m < n,
$$
we see that the Galois group ${\rm Gal}(K/L)$ is a subgroup of ${\rm Gal}(K/\mathbf{Q})$ of index less than $n$.  Since ${\rm Gal}(K/\mathbf{Q})\cong S_n$, and $S_n$ has no proper subgroups of index less than $n$, it follows that ${\rm Gal}(K/L)={\rm Gal}(K/\mathbf{Q})$, so that $L=\mathbf{Q}$.  Thus
$$
[K(\beta):\mathbf{Q}(\beta)] = \#{\rm Gal}(K(\beta)/\mathbf{Q}(\beta)) 
= \#{\rm Gal}(K/L) = [K:L] = [K:\mathbf{Q}],
$$
so that
$$
[K(\beta):K] = \frac{[K(\beta):\mathbf{Q}]}{[K:\mathbf{Q}]} = 
\frac{[K(\beta):\mathbf{Q}(\beta)]\cdot [\mathbf{Q}(\beta):\mathbf{Q}]}{[K:\mathbf{Q}]} = [\mathbf{Q}(\beta):\mathbf{Q}] = m.
$$
Therefore $[M(\beta):M]=m$ for every field $M$ with $\mathbf{Q}\subseteq M\subseteq K$, so in particular for $M=K(\alpha)$ we find that
$[\mathbf{Q}(\alpha,\beta):\mathbf{Q}(\alpha)]=m$, which means that $\mathbf{Q}(\alpha)$ and $\mathbf{Q}(\beta)$ are linearly disjoint over $\mathbf{Q}$.
If $m=n$ then this argument proves the result unless $[L:\mathbf{Q}]=n$.  Since the only index-$n$ subgroups of $S_n$ are the $n$ one-point stabilizers (assuming $n\ne 6$), it follows (for $n\ne 6$) that if $\mathbf{Q}(\alpha)$ and $\mathbf{Q}(\beta)$ are not linearly disjoint then $\mathbf{Q}(\alpha)$ and $\mathbf{Q}(\beta)$ are conjugate to one another over $\mathbf{Q}$.  Presumably there are quick arguments why this “almost surely” doesn't happen.
