Most likely value of rank if the entries of a matrix are randomly chosen from $[0,1]$ 
If the entries of a 3 by 3 matrix are chosen randomly between $0$ and $1$, what are the most likely dimensions of the four subspaces?

The solution to this is stated as most likey the matrix is $dim(C(A^T))=dim(C(A))=3$, see link
All possibilities $=2^9$
$\#$ favouring rank $3=2^3$
$\#$ favouring rank $1=2^2+2+1$
$\#$ favouring rank $2=2^3-(2^2+2+1)$
Looks like most likely the matrix can be of rank 2 ?
Update
Thanx @angryavian for the clarification, each entry of the matrix $x\in[0,1]$. I think in that case intuitively it seems colums of the matrix are most likely independent. But how do I mathematically prove that most likely the matrix is of full rank ?
Reference: 10, Problem Set 3.5, Dimension of Vector Spaces, Introduction to Linear Algebra - Ed 5, Gilbert Strang
 A: Suppose each column of the matrix is drawn independently from the uniform distribution on the cube $[0,1]^3$. By definition, the probability that each column lies in some region $A$ of the cube is equal to the volume of $A$. This is analogous to how the probability that a random real number (drawn from $[0,1]$) lies in some interval $[a,b] \subseteq [0,1]$ is equal to $b-a$, the length of the interval.
After drawing the first column, you need to draw the second column so that it does not lie on the subspace spanned by the first column (else the matrix will be rank-deficient). This subspace is a line inside the cube $[0,1]^3$ that has zero volume, so the probability the the second column lines in this subspace is zero. That is, the probability that the second column is linearly independent of the first column is $1$.
After drawing the first and second columns (where the second column is linearly independent of the first column), you need for the third column to lie outside the subspace spanned by the first two columns. This subspace is a plane inside the cube, and this plane also has zero volume, so again the probability that the third column lies in this subspace is zero. That is, the probability that the third column is linearly independent of the first two columns is $1$.

In general, this argument works for any distribution of $[0,1]^3$ that has a density function. For all such distributions, the probability of lying in a $1$- or $2$-dimensional subspace is zero, so the probability that a given column is linearly independent of the previous columns is $1$.
