Why is a singular matrix rare?

Thinking in terms of probability helps. If you have a continuous probability distribution defined on some space of matrices, then typically the singular matrices will have probability zero. Thinking in terms of the determinant: The determinant is a polynomial in the entries of the matrix. Setting it to zero gives a polynomial equation, which are defining (implicitely) some surface in the matrix space. This surface will have a reduced dimension , so its (Lebesgue) measure will be zero.

The number of $2\times2$ matrices over a field of $q$ elements is $q^4$.

The number of non-singular $2\times2$ matrices over a field of $q$ elements is $$(q^2-1)(q^2-q)=q^4-q^3-q^2+q$$ which means only $q^3+q^2-q$ out of $q^4$ are singular.

Here is an extension of Gerry's argument. There are $q^{n^2}$ matrices in $M_{n\times n}(\mathbb{F}_q)$ and $\prod_{k=0}^{n-1}(q^n-q^k)$ elements in $GL_n(\mathbb{F}_q)$.

$$\lim_{q\to\infty}\frac{\prod_{k=0}^{n-1}(q^n-q^k)}{q^{n^2}}=\lim_{q\to\infty}(q^{-n};q)_n=\lim_{q\to\infty}\prod_{k=0}^{n-1}\left(1-q^{-n+k}\right)=1$$ where $(q^{-n};q)_n$ is the $q$-Pochammer symbol.

Note that the fact that $\mathbb{F}_q$ has nonzero characteristic doesn't affect this argument, since the formula $\prod_{k=0}^{n-1}(q^n-q^k)$ is based on picking $n$ linearly independent vectors combinatorially from $(\mathbb{F}_q)^n$, a process which extends to the infinite case independently of characteristic.