Let $F = \mathbb{Z}_q$ be a finite field. Let $n \leq m' \leq m$. Pick a matrix $A \in F^{m \times n}$ such that each entry is chosen from a uniform distribution over $F$. Consider any $m' \times n$ submatrix of $A$ formed by choosing any $m'$ rows of $A$. What is the probability that every such submatrix $A'$ has full rank $n$?
Using the first equation here, I found that there are $\prod_{i=0}^{n-1} (q^{m'} - q^i) = q^{m' n} \prod_{i=0}^{n-1} (1 - q^{i-m'})$ matrices of size $m' \times n$ with rank $n$, so the probability that any arbitrary $A'$ has rank $n$ is given by $\prod_{i=0}^{n-1} (1 - q^{i-m'})$. I'm not sure if this helps to compute the probability that all $A'$ have full rank.
Even just a strong lower bound on this probability would help. Thank you!