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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!

eyehearyou
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  • Also see https://math.stackexchange.com/questions/1607395/probability-that-a-quadratic-equation-with-random-coefficients-has-real-roots, https://math.stackexchange.com/questions/624900/if-the-coefficients-of-the-quadratic-equation-ax2bx-c-are-u-i-i-d-ran-varia?noredirect=1&lq=1. – StubbornAtom Jun 15 '19 at 13:13
  • Thanks! I'll check 'em out. – eyehearyou Jun 15 '19 at 16:02

1 Answers1

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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 $c$...

gt6989b
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  • So I'm ok on the first part, the inner integral for $Pr(b>2\sqrt{ac})$, but then I think the outer double integral should be over ${a,c:ac<\frac{1}{4}}$. That would mean the integral would be something like $\int\int\limits_{{a,c: \text{ } ac<\frac{1}{4}}}Pr(b>2\sqrt{ac})dadc$. Does that make any sense? If so, then that's where I'm stuck... – eyehearyou Jun 14 '19 at 21:30