Find the probability that the equation $x^2+0.5x\sqrt{Y}+Z=0$ has real roots in x. Let Y and Z be independent random variables both of which have the exponential density $f(t)=0.25e^{-0.25t}, t>0$. Find the probability that the equation $x^2+0.5\sqrt{Y}+Z=0$ has real roots in x.
Evaluating the discriminant yields: $\frac{Y}{4} - 4Z\geq0$. Therefore, $Y\geq16Z$
Do I now just have to evaluate $\int^{\infty}_{0}\int^{16z}_{0} f(t)\,dxdy$ ?
 A: We want $\Pr(Y\ge 16Z)$. This is
$$\int_{z=0}^\infty \left(\int_{y=16z}^\infty (0.25)^2e^{-0.25 y}e^{-0.25z}\,dy     \right)\,dz.$$
A: Your integral is not the right one. First, a few reasons that have nothing to do specifically with probability:
First of all, you should not be integrating with respect to $x$ anywhere. The random variables are $Y$ and $Z$.
Second, since the inner integral has $z$ appearing in its limit, $z$ will appear as a variable in the outer integral. This hints that you should have $dy\,dz$ in that order. 
Third, since you're integrating with respect to $y$ and $z$, you don't want $t$ as your variable for $f$, because it's not going to do anything.
Fourth, the area of the $yz$-plane given by $y\geq16z>0$ gives you the integral limits $\int_0^\infty\int_{16z}^\infty dy\,dz$. Can you see why this describes an integral over all values where your inequality holds?
Then we come to the integrand. $Y$ and $Z$ are i.i.d. (independent identically distributed) random variables with pdf given by $f$. That means that their joint pdf is $g(y,z) =f(y)f(z)$, and that's the function you ought to integrate.
All in all we get that the integral you're supposed to calculate is
$$
\int_0^\infty\int_{16z}^\infty f(y)f(z)\:dy\,dz
$$
A: I'd go with the integral 
$$
\int_0^{\infty} \int_{16z}^{\infty} f(z)f(y) \, dy \, dz. 
$$
(The factor $f(z) f(y)$ corresponds to the join density of two independent i.i.d. random variables.) Can you evaluate?
Hope that helps,
