Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A, B, and C are random variables uniformly distributed over (0,1). 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
$$\begin{align}
\int_0^1 \int_0^1 \int_{\min\{1, \sqrt{4ac}\}}^1 1 \;\text{d}b\,\text{d}c\,\text{d}
&a= \int_0^1 \int_0^{\min(1, \frac{1}{4a})}\int_{\sqrt{4ac}}^1  1\;\text{d}b\,\text{d}c\,\text{d}a\\
&= \int_0^{\frac{1}{4}} \int_0^1 \int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a + \int_{\frac{1}{4}}^1 \int_0^{\frac{1}{4a}}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a
\end{align}$$
why the middle integrate from $0$ to $\min(1, \frac{1}{4a})$ from the second integral...where does $\frac{1}{4a}$ 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 
 A: To resolve $\min\{1,\sqrt{4ac}\}$, we need to figure out which of the two arguments is smaller. If it's $1$, the integral is $0$ and thus doesn't contribute. For it to be $\sqrt{4ac}$, we need to have $\sqrt{4ac}\le1$, and thus $c\le1/(4a)$. This is the answer to your question where $1/(4a)$ comes from.
I'm not sure how to answer your question why the $\min$ doesn't go to the front integral. My counter-question would be how you'd propose to move it to the front integral such that the result is equivalent. Unless you can come up with such a proposal, I'd suggest to concentrate on why the form written here is equivalent to the one in the previous step. Generally speaking, it makes sense that resolving a minimum in a limit of the innermost integral affects the limits of the immediately enclosing integral and not of a more remote enclosing integral.
The breaking up into two integrals is again the direct result of resolving the minimum in the upper limit of the second integral. That minimum is $1$ if $1\le1/(4a)$, i.e. if $a\le\frac14$, and is $1/(4a)$ otherwise; thus we have to split the outer integral over $a$ into two parts according as $a\lessgtr\frac14$.
