Probability that a determinant is equal to zero. Let $A \in M_{n,n}(\mathbb{R})$. 
What is the probability that, when I fill in the entries with random real numbers, $det(A) =0$? This looks like an interesting problem, as having determinant zero is equivalent to a lot of other things. For example, the answer to this question gives us information about the percentage of systems $AX = B$ where $A \in M_{n,n}(\mathbb{R})$ that have a unique solution (Cramer's rule)
And what if:
1)  $A \in M_{n,n}(\mathbb{C})$? 
2)  All entries must be natural numbers/integers/rational numbers
 A: Even in the simplest case of $n=2$ over $\mathbb{R}$ you are asking what is the probability that $ad = bc$ where $a,b,c,d$ are random. This is clearly 0.
All others can be argued similarly. To make this interesting, ensure you are sampling entries from a finite distribution.
A: For part $2)$ - Firstly note that the determinant of a 'rational' matrix $A$ is $0$ if and only if the determinant of the matrix $\lambda A$ is $0$, where $\lambda$ is the product of all the denominators of the entries of $A$. Clearly $\lambda A$ has integer entries. Therefore if we can show that the probability of picking a matrix with integer coefficients that has determinant $0$ is $0$, then the same holds for rational matrices.
By symmetry, given $n \in \mathbb{N}$, the probability that a randomly selected integer matrix has determinant equivalent to $0$ modulo $n$ is $1/n$. The probability that a random integer matrix has determinant $0$ is less than this. Hence the probability is less than $1/n$. Since this holds for all $n$, letting $n$ tend to infinity shows that this probability is $0$. Hope this helps.  
