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.
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 $$ p+(1-p)q = {\small{\frac{29}{216}}}+{\small{\frac{11}{72}}}\ln(2) \approx .2401567452 $$