Suppose I have a $m \times n$ ($m>>n$) zero matrix (all of the elements are $0$). I want to flip $k$ ($k \ge n$ and can be controlled by the user) elements of the matrix randomly. After this trick, what is the probability of the matrix is full column rank?
Take $3 \times 2$ matrix $A$ as an example. If I flip $3$ ($k=3$) elements out of $6$ elements, then
$\text{Prob} (\text{A is full column rank})$ = $C_3^2 \times 2! \times \dfrac{C_4^1}{C_6^3} =\dfrac{1}{5}$.
Example
$$\begin{pmatrix} 0&0 \\ 0&0 \\ 0&0 \end{pmatrix} \to \begin{pmatrix} 1&1 \\ 0&1 \\ 0&0 \end{pmatrix} $$ If $k=4$, the following case means the matrix is not full column matrix.
$$\begin{pmatrix} 0&0 \\ 0&0 \\ 0&0 \end{pmatrix} \to \begin{pmatrix} 1&1 \\ 1&1 \\ 0&0 \end{pmatrix} $$ Is there any general way to calculate the probability when the matrix is of full column rank? Or the parameter $k$ can be optimized to make the matrix be full column rank?