Suppose X has the uniform distribution on [0,1]. Suppose that given X, Y has the Gamma distribution with mean X and variance 3X. Suppose $X$ has the uniform distribution on $[0,1]$. Suppose that given $X$, $\,Y$ has the Gamma distribution with mean $X$ and variance $3X$.
a) Find $E(Y)$
b) Find $Var(Y)$
c) Give and integral expression for $P(XY>1)$ without solving it.
This is what I think I should do.
for parts (a), $E(Y) = E(E(Y\mid X))$. The expectation of $Y$ given $X$ is $X$. So,
$E(Y) = E(E(Y|X)) = \int_0^1 X \times1 \ dX = \frac{1}{2}$
Also, for part(b),
$Var(Y) = E(Var(Y\mid X)) + Var(E(Y\mid X)) = E(3X) + Var(X) = 3E(X) + \frac{1}{12} = 3 \times 0.5 + \frac{1}{12} = \frac{19}{12}$
if part (a) and (b) are correct, how should I continue for part (c)?
Thanks.

Your answers for $(a)$ and $(b)$ are correct.  For $(c)$, we want $\Pr[XY > 1]$, and the most natural way to do this is to condition on $X$:  $$\Pr[XY > 1] = \int_{x=0}^1 \Pr[Y > 1/X \mid X = x]f_X(x) \, dx = \int_{x=0}^1 \int_{y=1/x}^\infty \frac{y^{x/3-1} e^{-y/3}}{3^{x/3} \Gamma(x/3)} \, dy \, dx.$$
Here I have used the fact that if the conditional mean and variance are $X$ and $3X$, then this implies the gamma distribution must have shape $X/3$ and rate $1/3$.
