Probability that $\text{det}(A)$ is an even number.

Let $A=\left(\matrix{a&b\\c&d}\right)\in\text{Mat}_2(\mathbb{Z})$. $\text{det}(A)=2n$ for some $n\in\mathbb{Z}$ iff $\text{det}(A')=0$, with $$ A'=\left(\matrix{a_2&b_2\\c_2&d_2}\right)\in\text{Mat}_2(\mathbb{Z}_2), $$ where the coefficients are reduced modulo 2.

The matrices with zero determinant in $\text{Mat}_2(\mathbb{Z}_2)$ are: \begin{align} \left(\matrix{0&0\\0&0}\right),&&\left(\matrix{0&1\\0&0}\right),&&\left(\matrix{1&0\\0&0}\right),&&\left(\matrix{0&0\\1&0}\right),&&\left(\matrix{0&0\\0&1}\right),\\ \left(\matrix{1&1\\0&0}\right),&&\left(\matrix{1&0\\1&0}\right),&&\left(\matrix{0&1\\0&1}\right),&&\left(\matrix{0&0\\1&1}\right),&&\left(\matrix{1&1\\1&1}\right).\\ \end{align}

So the required probability is $$ \frac{10}{|\text{Mat}_2(\mathbb{Z}_2)|}=\frac{10}{|\mathbb{Z}_2|^{2\cdot 2}}=\frac{10}{16}=\frac{5}{8}. $$

Assuming that we are dealing with simple random samples, since the determinant of the matrix $ad-bc$ is even if $ad$ and $bc$ are both even or both odd, we obtain

$$P=\overbrace{\left(\frac14\right)^2}^{\color{blue}{\text{both odd}}}+\overbrace{\left(1-\left(\frac12\right)^2\right)^2}^{\color{red}{\text{both even}}}=\frac1{16}+\frac9{16}=\frac 5 8$$

Otherwise, as noticed by David C. Ullrich, we need to specify a probability distribution for $A$.

There's no such thing as a "random matrix". Or rather, there are infinitely many different things that the phrase "random matrix" might mean - before you can answer any questions about the probability of this or that given a random matrix you need to specify a probability distribution for $A$.

The two other answers both entail (reasonable) unstated assumptions about this distribution. For a detailed discussion of a similarly poorly-posed problem see Bertrand's paradox.