Let be
a $n \times n$ Symmetric Matrix over Finite Field with $q$ elements, where the green color represents 0's and the black color non-zero entries.
How I will be able to demonstrate that probability of the rank should be $2s$ is $1-(1-1/q)\times\cdots\times(1-1/q^s)$?
EDIT Sorry I think that my question is wrong, because I test using python program that formula and that is wrong then I want re-formulate the question: What's the probability of that matrix to have rank $2s$?
First consider the rows below s.
It is almost obvious that maximum rank of the mentioned matrix is $2s$, since there can be at most $s$ independent rows below $s$th row and the first $s$ rows will only add $s$ to the rank of $A$.
the probability of the last $n-s$ rows having rank $s$ should be computed (which is equal to the probability of $(n-s) \times s$ matrix having full rank), and then knowing it spans the space, it can cancel out the $s \times s$ square. The probability of the first $s$ rows adding $s$ to the rank is also 1, since the matrix is symtetric and the $n \times s$ matrix that needs to be full rank is the transpose of previous.
for such matrix, looking now columnwise we need $s$ independent columns in the $n-s$ dimensional space.
Assuming uniform distribution. For each row not being in the subspace of the previous rows, we have a probability of $1-p^{i-1}$. So, $p=1/q$ is the probability of each symbol
$(1-p^0)(1-p^1)(1-p^2)...(1-p^{s-1})$
so in conclusion the probability of $A$ being of rank $2s$ is $\prod_{i=0}^{s-1}(1-p^i)$.