Let $\mathbb{F}_{11}=\{\overline{0},\overline{1},\ldots,\overline{10}\} $ be a finite field and $A=(a_{ij})_{3 \times 3}$ a matrix such that $a_{ij} \in \mathbb{F}_{11}, \ \ \forall \ 1\leq i,j\leq 3$. What is the probability that $rank(A)=1$.
I tried as follows: There is $11^9$ matrices of the described form. If $v_1, v_2, v_3 \in \mathbb{F}_{11}^3$ are the row vectors if $A$, we need that $dimV=1$, where $V=span\{v_1,v_2,v_3\}$. So we can choose the vector $v_1$ of $11^3-1$ different ways ($v_1\neq 0$).Now, there are $11$ ways for choose teh vector $v_2$ and $11.11=11^2$ ways for choose the vector $v_3$...Is this idea correct?
Your counting is not entirely correct, as the first row can also be zero (as already commented by user1551). But the basic idea works and refining your argument, we get a correct solution.
Solution. If the first row is non-zero ($11^3 - 1$ possibilities), there are $11^2$ possibilities for the other two rows. (The only condition is that they are contained in the span of the first row, which is of size $11$.) If the first row equals zero and the second row is non-zero ($11^3 - 1$ possibilities), there are $11$ possiblilities for the last row. If the first two rows are zero, the last row must be nonzero, so $11^3 - 1$ possibilities. So in total there are $$ (11^3 - 1) \cdot 11^2 + (11^3 - 1) \cdot 11 + (11^3 - 1) = 176890 $$ rank $1$ matrices, and hence the requested probability equals $$ \frac{176890}{11^9} \approx 0.00750186\%. $$
Alternative 1. Knowing a bit more about the combinatorics of finite vector space, you can do it in this way for $n\times n$ matrices of rank $1$ over a general finite field $\mathbb{F}_q$: The rank of a matrix is the dimension of its row space. So the number of possible row spaces equals the number $\frac{q^n - 1}{q-1}$ of $1$-dimensional subspaces $W$ of $\mathbb{F}_{q}^n$. For fixed $W$ the number of $n$-tuples of vectors spanning $W$ equals $q^n - 1$. (Note that $\#W = q$, and the single $n$-tuple $(0,0,0)$ does not span $W$, but all other $n$-tuples over $W$ do.)
Hence the number of rank $1$ matrices of size $n\times n$ over $\mathbb{F}_{q}$ is $$ \frac{q^n - 1}{q-1} \cdot (q^n - 1) = \frac{(q^n - 1)^2}{q-1}\text{.} $$
Alternative 2. There is another alternative if you know that a rank $1$ square matrix has always the form $v\cdot w^{\top}$ with non-zero vectors $v,w$. One checks that two such matrices $v\cdot w^{\top}$ and $v' \cdot (w')^\top$ are equal if and only if $v' = \lambda v$ and $w' = \lambda^{-1} w$ with a non-zero scalar $\lambda$. Hence each rank $1$ matrix has $q-1$ representations $v\cdot w^{\top}$, and again we get their number as $\frac{(q^n - 1)^2}{q-1}$.