Let $A$, $B$, and $C$ be independent random variables, uniformly distributed over $[0,9], [0,3]$ and $[0,5]$ respectively. What is the probability that both roots of the equation $Ax^2+Bx+C=0$ are real?
I know I need $B^2>4AC$ but not sure how to get probability.
This is more of a rough sketch: With a slight abuse of notation:
$|A| = 9-0 = 9$
$|B| = 3-0 = 3$
$|C| = 5-0 = 5$
First the probability distribution of $B^2$:
$P(B^2 < x) = P(-\sqrt{x} < B < \sqrt{x}) = P(0 < B < \sqrt{x}) = \frac{\sqrt{x}}{|B|}$
$\implies f_{B^2}(x) = \frac{1}{2\sqrt{x}\ |B|}$
Now consider: \begin{align*} P(B^2-4AC > 0) &= P(C<\frac{B^2}{4A})\\ &=P(C < \frac{B^2}{4A}, \frac{B^2}{4A} > |C|) + P(C < \frac{B^2}{4A}, \frac{B^2}{4A} < |C|)\\ &=P(C < |C|, \frac{B^2}{4A} > |C|) + P(C < \frac{B^2}{4A}, \frac{B^2}{4A} < |C|)\\ &= P( C < |C| | \frac{B^2}{4A} > |C|) \times P(\frac{B^2}{4A} > |C|) + P (C < \frac{B^2}{4A}) \times P(\frac{B^2}{4A} < |C|) \\ &= 1 \times P({B^2} > 4A|C|) + \int \int \frac{B^2}{4A|C|} \times P(B^2 < 4A|C|) dA dB \end{align*}
Let $B'=B^2$ \begin{align*} P(B^2 < 4A|C|) &= \int_0^{|B'|} \int_{\frac{B'}{4|C|}}^{|A|} \frac{\sqrt{4|C|}}{|B'|} \sqrt{A}\ dA\ dB'\\ &= \frac{\sqrt{4|C|}}{|B'|} \int_0^{|B'|} \frac{2}{3} A^{3/2}\big|_{\frac{B'}{4|C|}}^{|A|} dB'\\ &= \frac{2}{3}\frac{\sqrt{4|C|}}{|B'|} \big(|A|^{3/2}B' - \frac{2}{5}\frac{B'^{5/2}}{(4|C|)^{3/2}} \big)\big|_{0}^{|B'|}\\ &= \frac{2}{3}\frac{\sqrt{4|C|}}{|B'|} \big(|A|^{3/2}|B'| - \frac{2}{5}\frac{|B'|^{5/2}}{8|C|^{3/2}} \big)\\ \end{align*}
\begin{align*} \int \int \frac{B^2}{4A|C|} \times P(B^2 < 4A|C|) &= \frac{\sqrt{4|C|}}{4|C||B'|} \int_0^{|B'|} \int_{\frac{B'}{4|C|}}^{|A|} \frac{B'}{A} \sqrt{A} \ dA \ dB'\\ &= \frac{\sqrt{4|C|}}{4|C||B'|} \int_0^{|B'|} {2B'}\sqrt{A}\big|_{\frac{B'}{4|C|}}^{|A|} \ dB'\\ &= \frac{\sqrt{4|C|}}{2|C||B'|} \int_0^{|B'|} B'(\sqrt{|A|}-\sqrt{\frac{B'}{4|C|}}) \ dB'\\ &= \frac{\sqrt{4|C|}}{2|C||B'|} \big( \sqrt{|A|} \frac{|B'|^2}{2} - \frac{2}{5} \frac{|B'|^{5/2}}{\sqrt{4|C|}} \big) \end{align*}
