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?



Strictly positive solutions of a random linear system
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 $m\to\infty$.


