How do you handle this kind of probability? What is the probability of selecting a singular matrix from $\Bbb{R}^{3\times 3}$? 
I calculated it to be zero based on their being approximately $9$ degrees of freedom to choose entries of $A$ such that $\det(A) \neq 0$, and $9$ degrees of freedom in choosing any $A \in \Bbb{R}^{3\times 3}$, and took $9/9 = 1$ as the probability of selecting a non-singular.  However I approximated the first number in the sense that if a variable (degree of freedom) could be chosen from $\Bbb{R} - \{0\}$ then it counts as $1$ full degree of freedom the same as a choice from $\Bbb{R}$ untouched.  So this obviously doesn't work??
Or does it work since singular matrices are extremely rare but we only look at matrices for relatively small (human-sized) values?  I can't explain it.  How would you go about calculating this type of probability?  Thanks.
 A: As the comments have noted, your question is difficult to answer because we don't know how to choose a random real number.  In the language of probability theory, the Lebesgue measure of $\mathbb{R}$ is infinite, and probability studies finite measure spaces (so $\mathbb{R}$ under Lebesgue measure is excluded).
However, we can still ask the following measure-theoretic question:

In the infinite measure space $M_3(\mathbb{R})\cong\mathbb{R}^9$, what is the measure of the set of singular matrices?

The answer to this question is indeed $0$, which validates your intuition that "most" matrices are invertible.  The singular matrices have measure $0$ because they are a hypersurface, that is, they are defined by the zero set of a single polynomial equation (the determinant).  In general, proper Zariski closed sets have measure $0$.
Edit: As requested by Rookatu's comment, here is a bit more, although I'm not sure if you have any background in Real Analysis, so this may not be helpful to you at the moment.
Suppose that $\mu$ is a probability measure on the Borel subsets of $\mathbb{R}^9$ that is absolutely continuous with respect to Lebesgue measure $m$.  This means that $m(A)=0$ implies $\mu(A)=0$ for all measurable sets $A$.  This is equivalent to the existence of some Lebesgue integrable function $f:\mathbb{R}^9\to\mathbb{R}$ such that
$$\mu(A)=\int_Af\;dm$$
for all measurable $A$.  This function $f$ is called the density of $\mu$, and in our case, it is a probability density function, that is, its integral over all of $\mathbb{R}^9$ is $1$.
Now, as noted above, if $A$ is the set of singular matrices in $\mathbb{R}^9$, then $m(A)=0$, so by absolute continuity, we also have $\mu(A)=0$.  All of this shows that if you describe your process of picking a matrix at random using a probability density function $f:\mathbb{R}^9\to\mathbb{R}$, no matter which $f$ you choose, the answer to your problem is the same: the probability that a random matrix is singular is $0$.
