linear dependncy of a random vector with respect to a reduced row echelon form in a finite field

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Given a matrix with elements from a finite field $\mathbb{F}_q$, $A\in\mathbb{F}_q^{N\times M}$, where $q$ is the size of the field, $N<M$. Suppose that $A$ in the reduced row echelon form. Obviously, the rank of such $A$ is $N$.

Suppose that now I generate a $M$-dimensional random vector $\mathbf{f}\in\mathbb{F}^{1\times M}$, with elements uniformly randomly chosen from $\mathbb{F}_q$. Is that possible to exactly compute the probability that $\mathbf{f}$ is linearly independent with those $N$ rows of $A$? I feel like it should be close to $1-\frac{q^N}{q^M}$, but not much sure. If it is not, any clue to look into the problem? Thanks.

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Yes, it is possible to exactly compute, and we need only classical probability. Your question amounts to choosing an $N$-dimensional subspace of $\mathbb{F}_q^M$ (the row space of your given matrix which you said has rank $N$) and then asking the probability that a randomly chosen vector does not lie in that subspace. Well since the subspace contains $q^N$ elements and the whole space contains $q^M$ elements, and each vector is equally likely to be selected, the probability is $$ \frac{q^M -q^N}{q^M} = 1 - \frac{q^N}{q^M} $$ as you suspect.