Strictly positive solutions of a random linear system Suppose $B\in\mathbb{R}^{m\times n}$ is a random binary matrix with i.i.d entries and $c\in \mathbb{R}^m$ is a strictly positive vector, that is $c_i>0$ for $i=1,2,\cdots m$. Also assume $m<n$, basically meaning that $B$ is a fat matrix. Is there any result (or any suggestion on how to approach the problem) that states for sufficiently large $n$, the linear system 
$Bx=c$
has a strictly positive solution almost surely (obviously among the many possible ones, there is one would this property). Basically I am looking for a tail bound like
$\mathbb{P}(\nexists x>0: Bx=c)\leq f(n,m)$
where $f(n,m)\to 0$ as $n \to \infty$ and $m$ stays fixed. Any suggestions on finding a tail bound like above?
 A: Let $1\leq i\leq m$. The probability that $B$ has no column with a unique one in the $i$th position is $(1-2^{-m})^n$. Thus the probability that $B$ has each such column is at least $1-m(1-2^{-m})^n$. If $B$ has all such columns (let their numbers be $j_1,\dots,j_m$ respectively) then one may assign very small values to $x_k$'s with $k\notin\{j_1,\dots,j_m\}$, and then determine $x_{j_i}$'s so that $Bx=c$. Thus one may take 
$$
  f(m,n)=m(1-2^{-m})^n.
$$
This is small provided that $2^m\log m=o(n)$.
ADDENDUM. On the other hand, let us see that no essentially better uniform bound is possible; that is, there exist suitable columns $c$ such that the probability is not that lagre. 
Let $c_m>c_1+c_2+\dots+c_{m-1}$. Then a positive solution is possible only if some column in $B$ is equal to $e_m$; otherwise the sum of the first $m-1$ coordinates in $Bx$ majorizes the last one.
Thus we have for this case $f(n,m)\geq (1-2^{-m})^n$ which tends to a nonzero limit as $n=2^m$ and $m\to\infty$.
A: A very related study in the case of gaussian entries is the paper :
"Positive solutions for large random linear systems"
https://arxiv.org/abs/1904.04559
