Q:

Solving probability of inequalities with random variables

Given an inequality (ex: $b^2-4ac>0$) and distributions of the variables (ex: a,b,c all fall between 0,1) how would one go about solving for the probability that this is true.
I really gave it an honest effort and got somewhere.  I determined for something simple such as $a^2-b>0$ I could simply integrate both from between their respective range (i.e. 0-1) and take the difference (correct me if I am wrong, but what graphs had me thinking.  For my little quadratic formula problem I tried to interpret it also as a graph but visualizing the overlap of 3D space is slightly beyond my cognitive capacity :P.  I am looking for a (slightly) foolproof way to approach this variant of problem, and appreciate any help I can get.
NOTE: This is not homework of any sorts, and I apologize if my question is slightly unintelligible.  I am simply an ambitious high-school student curious on how one might solve this.

A:

Assume $0 \le a,b,c \le 1$.

If $b^2 > 4a$, then $b^2 > 4ac$ is automatic.

Let $p$ be the probability that $b^2 > 4a$.$\;$Then
$$p=\int_{0}^1\int_{0}^{{\large{\frac{b^2}{4}}}} 1\;da\,db = \frac{1}{12}$$
Let $q$ be the probability that $b^2 > 4ac$ given that $b^2 < 4a$.$\;$Then
$$
q=
\int_0^1
\int_{\large{\frac{b^2}{4}}}^1
\int_0^{\large{\frac{b^2}{4a}}} 
1
\;dc\,da\,db 
= 
{\small{\frac{1}{18}}}+{\small{\frac{1}{6}}}\ln(2)
$$
Hence, the probability that $b^2 > 4ac$ is equal to
$$
p+(1-p)q 
=
{\small{\frac{29}{216}}}+{\small{\frac{11}{72}}}\ln(2)
\approx 
.2401567452
$$

