My Math Forum Understanding the setup for the probability that $Ax^2+Bx+C$

 March 1st, 2014, 08:41 PM #1 Newbie   Joined: Mar 2014 Posts: 1 Thanks: 0 Understanding the setup for the probability that $Ax^2+Bx+C$ Suppose that $A, B,$ and $C$ are independent random variables, each being uniformly distributed over $(0,1)$. What is the probability that $Ax^2 + Bx + C$ has real roots? First, I set $P(B^2 - 4AC \ge 0)$ Then I am told that \begin{align} \int_0^1 \int_0^1 \int_{\min\{1, \sqrt{4ac}\}}^1 1 \;\text{d}b\,\text{d}c\,\text{d} &a= \int_0^1 \int_0^{\min\{1, 1/4a\}}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a\\ &= \int_0^{1/4} \int_0^1 \int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a + \int_{1/4}^1 \int_0^{1/4a}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a \end{align} why the middle integrate from 0 to min{1, 1/4a} from the second integral...where does 1/4a come from? why the min{...} does not go to the front integral? why they break up into last step like this (I refer to one integral + another integral) ? Thanks a lot
 March 2nd, 2014, 12:40 PM #2 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 Re: Understanding the setup for the probability that $Ax^2+B Something is wrong with your tex.  Tags$ax2, probability, setup, understanding

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