Random variables $X$, $Y$, $Z$, probability of real roots of $Xt^2+Yt+Z=0$

Question

The random variables $X$, $Y$, $Z$ are independent and uniformly distributed in $[0,1]$. What is the probability of real roots of $Xt^2+Yt+Z=0$?

So the equation has a real root when $Y^2\geq 4XZ$. Given that the variables are independent, the joint density is $f(x,y,z) = 1$ in the unit cube $[0,1]^3$, and $0$ otherwise. My guess is that I need to integrate the volume between the surface $y^2=4xz$ and $y=0$ inside the unit cube. My calculus is rusty so I need help on how to proceed.

Answer 1

Hint: If you sketch the region, you should be able to see that the integral is $$ \int_0^1 \int_0^{\min\{1, 1/4z\}} \int_{2\sqrt{xz}}^1 1 \; dy \; dx \;dz $$$$ = \int_0^{1/4} \int_0^{1} \int_{2\sqrt{xz}}^1 1 \; dy \; dx \;dz + \int_{1/4}^{1} \int_0^{1/4z} \int_{2\sqrt{xz}}^1 1 \; dy \; dx \;dz. $$