Probability of having real roots Let $U,V,W$ are independent random variables with $\mathrm{Uniform}(0,1)$ distribution. I am trying to find the probability that $Ux^{2}+Vx+W$ has real roots, that is, $P(V^{2}-4UW> 0)$
I have solved this question using double integral but how to do this using triple integral.
My Approach:
I started with cdf:
$P(V^{2}-4UW >0) =P(V^{2} > 4UW) = P(V>2\sqrt{UW})$ = $\int\int_{2\sqrt{uw}}^1 P(V>2\sqrt{UW}) dU dW$
=$\int\int\int_{2\sqrt{uw}}^1 vdU dW dV$
I am finding hard time to get the limits of integral over the region in 3 dimensions.
Using double integral:
$P(V^{2}-4UW >0) =P(V^{2} > 4UW) = P(-2\ln V <-\ln 4 - \ln U - \ln W) = P(X <-\ln 4 +Y)$
where $X=-2 \ln V, Y = - \ln U -\ln W $
$X$ has $\exp(1)$ and $Y$ has $\mathrm{gamma}(2,1)$ distribution.
$P(X <-\ln 4 +Y) = \int_{\ln4}^\infty P(X < -\ln 4 +Y) f_Y(y) dy $
$$=\int_{\ln 4}^\infty\int_0^{-\ln 4+y} \frac{1}{2} e^{-\frac{x}{2}}ye^{-y} dxdy $$
Solving this I got $0.2545$.
Thanks!
 A: Here is a solution without multiple integrals calculation (because I don't like multiple integrals). Actually it only uses three elementary simple integrals. 
$$
P(V^{2}-4UW \leq  0) = E\bigl[P(V^{2}-4UW \leq 0 \mid U,W)\bigr] = E\bigl[f(U,W)\bigr]$$ where $f(u,w)=P(V^{2}-4uw \leq 0)= \min\bigl\{1, 2\sqrt{uw}\bigr\}$.
$$
E\bigl[f(U,W)\bigr] = E[g(W)]
$$
where 
$$\begin{align}
g(w) & = E\bigl[\min\bigl\{1, 2\sqrt{Uw}\bigr\}\bigr] 
= 1 \times \Pr(2\sqrt{Uw}>1) + E\bigl[2\sqrt{Uw} \mathbf{1}_{2\sqrt{Uw}\leq 1}\bigr] \\
& = \Pr(U>\frac{1}{4w}) + 2\sqrt{w}E\bigl[\sqrt{U} \mathbf{1}_{U \leq \frac{1}{4w}}\bigr]  \\
& = \max\bigl\{0, 1 - \frac{1}{4w}\bigr\} +  2\sqrt{w} \times \frac{2}{3} \times \min\bigl\{1, \frac{1}{{(4w)}^{\frac{3}{2}}}\bigr\} \\ 
& =\begin{cases} 
 0 + \frac{4}{3}\sqrt{w}  & \text{if } w \leq \frac{1}{4} \\
1 - \frac{1}{4w} + \frac{1}{6w}  & \text{if } w > \frac{1}{4}
\end{cases}, \end{align}$$
and we get 
$$ E[g(W)] = \frac{1}{9} + \frac{3}{4} - \frac{1}{12} \log 4 = \frac{31}{36}-\frac{\log 2}{6},$$
and finally 
$$P(V^{2}-4UW >  0) = \frac{5}{36} + \frac{\log 2}{6} \approx 0.2544134.$$
A: By drawing the regions and taking integrals we get:
$P(V^{2} - UW \geq 0) = P(V^{2} \leq  UW, 0\leq U,V,W \leq 1)$
$=P(0 \leq V \leq 2\sqrt{UW},0\leq U,V,W \leq 1)$
$=P(0 \leq V \leq min(1,2\sqrt{UW}),0\leq U,V,W \leq 1)$
$=\int_0^\frac{1}{4}\int_0^1\int_0^{2\sqrt{uw}}dudvdw + \int_\frac{1}{4}^1\int_0^{\frac{1}{4u}}\int_0^{2\sqrt{uw}}dudvdw + \int_{\frac{1}{4}}^1\int_\frac{1}{4u}^1\int_0^1dudvdw $
Solving the integrals we get the same answer, that is , approx 0.2544.
