Probability that $AX^2+BX+C=0$ has only real roots for $A,B,C \sim Unif(0,1)$

Question

From chapter 6, practice exercise 26 (b) from "A First Course in Probability" by Sheldon Ross (working it for my own recreation):

Given R.V.s $A,B,C\sim_{iid}Unif(0,1)$ and asked to find the probability that $AX^2+BX+C=0$ has real roots (ie $B^2>4AC$).

Can someone help me understand how to set up the integral to compute the desired probability: $$ Pr(B^2>4AC)=Pr(B>2\sqrt{AC}) $$ *$\{B<-2\sqrt{AC}\}$ has probability zero since $B\sim Unif(0,1)$ and, therefore, cannot be negative.

I understand that, in order for $B$ to be between 0 and 1, we need $0<\sqrt{AC}<\frac{1}{2} \rightarrow 0<AC <\frac{1}{4}$, but I am having trouble coming up with an integral that makes sense.

Thanks!

Answer 1

So you need to evaluate $$ \begin{split} I &= \int_0^1 \int_0^1 \int_0^1 \mathbb{I}_{(b^2 > 4ac)}\ da\ db\ dc \\ &= \int_0^1 \int_0^1 \int_0^1 \mathbb{I}_{(b > 2\sqrt{ac})}\ da\ db\ dc \\ &= \int_0^1 \int_0^1 \mathbb{P}\left[b > 2\sqrt{ac}\right]\ da\ dc \end{split} $$ if you integrate $db$ first. Given $(a,c) \in \mathbb{R}^2$, can you compute this probability, assuming $A,B,C$ are independent? Then integrate wrt $a$ and $c$...