Probability of real roots Let's take random $a, b \in [0,2] $.
What is a probability that equation $ax^2+bx+1=0$ has two real roots?
My solution:
$\Delta = b^2-4a>0$
$a< \frac{b^2}{4}$
$P( \Omega) = 2 *2=4$
$\mathrm{P}(A) = \int^2_0 \frac{b^2}{4}\mathrm{d}b$
and the answer is $\frac{\mathrm{P}(A)}{\mathrm{P}(\Omega)}$
Is it correct?

It might be a bit more transparent to convert to the Uniform(0, 1) distribution rather than use Uniform(0, 2). Given the setup, we have 
$P\big(a < \frac{b^2}{4}\big) \Longleftrightarrow P\big(a^* < \frac{b^{*2}}{2}\big)$
where $a^* = \frac{a}{2}$ and $b^* = \frac{b}{2}$ are from independent Uniform(0, 1) distributions. 
Then $P\big(a^* < \frac{b^{*2}}{2}\big)$ is simply the proportion of the unit square for which
$a^* < \frac{b^{*2}}{2}$ which is
$\int_0^1 \frac{b^{*2}}{2}db^*$ = $[\frac{1}{6}b^{*3}]_0^1$ = $\frac{1}{6}$
(in agreement with your result).
