Assume the following random matrix with $N$ rows and $L$ columns consisting of elements in {0,1}. \begin{bmatrix} 1 &0 & 1 & ...&0 \\ 1 & 0 & 0& ... &1 \\ 1 & 0 & 1 & ... & 0\\ \vdots& \vdots&\vdots&\ddots& \vdots\\ 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 1 \end{bmatrix} The probability of an element being a $1$, (independent of the matrix elements) is $\epsilon$. What is the probability of at most $K$ elements equal to $1$ in a column?
My own calculation is $P=\sum_{k=0}^{K}\begin{pmatrix}N\\ k\end{pmatrix}\epsilon^k(1-\epsilon)^{(N-k)}$, but I am not sure about it since does not depend on $L$.
Verification of the above equation, an approximate, or an educated guess, all would be useful.
First, let $N=L=3$ and $K=2$.
The following matrix satisfies that there are at most two rows having at least one $1$'s:
$$ \left[ \begin{array}{c|c|c} 1&1&1\\ \hline 0&0&1\\ \hline 0&0&0 \end{array} \right] $$
It follows that every column contains at most two $1$'s
In the following matrix there are three rows containing at most one $1$'s, yet none of the columns contain more than two $1$'s.
$$ \left[ \begin{array}{c|c|c} 1&1&1\\ \hline 1&0&0\\ \hline 0&0&1 \end{array} \right] $$
In general the probability that one given column contains at most $K$ $1$'s is - as you wrote: $$P=\sum_{k=0}^{K}\begin{pmatrix}N\\ k\end{pmatrix}\epsilon^k(1-\epsilon)^{(N-k)}.$$
Since the columns are independent, the answer to the OQ is $P^L$.