Probability of the average lifetime of $n$ machines 
Consider $n=35$ independent machines, each with a lifetime of $V=X+Y$ :


$X \sim U(0,11)$


$Y|X=x \sim Exp(\frac{1}{x})$


It is unknown whether $X$ and $Y$ are independent.


How can I determine the probability of the average lifetime of those $n$ machines to be higher than $k=10$ ?

Attempt:
I was unable to determine the distribution of $Y$, but I've determined the conjoint distribution: $f(x,y)=\frac{1}{11xe^{(y/x)}}$
The average lifetime value of our $n$ machines is:
$\overline V=\frac{1}{n}\sum\limits_{i=1}^n V_i$
I want to find:
\begin{equation} \mathbb P(\overline V>k)=1- \mathbb P(\overline V \leq k)\end{equation}
\begin{equation} =1- \mathbb P(\sum\limits_{i=1}^n V_i \leq n*k) \end{equation}
But I'm not sure what to do from here. Could I use the Central Limit Theorem or am I way off track ?
Any help is appreciated
 A: Here's a sketch of the derivation. I think the distribution of $X$ should be $R[\varepsilon, 11]$ because the parameter of the exponential distribution can't be undefined.
First, derive the joint distribution, $f_{XY}(xy) = f_{Y|X}(y|x) \cdot f_X(x)$. Once you've done that, derive the expectation of $XY$, you will need it later.
$$
\mathbf{E}XY = \int_{\varepsilon}^{11}\int_{0}^{\infty}xyf(xy)dydx
$$
Derive marginal expectation of $Y$:
$$
\mathbf{E}Y = \int_{\varepsilon}^{11}\mathbf{E}[Y|X]f_X(x)dx = \frac{1}{11} \int_{\varepsilon}^{11}xdx
$$
This is because Exponential mean and variance are $\frac{1}{\lambda}= x$. Now, you can get the mean of $V$, by linearity:
$$
\mathbf{E}V = \mathbf{E}X + \mathbf{E}Y
$$
To get the variance of V,
$$
Var[V] = Var[X] + Var[Y] + 2 Cov(X,Y)
$$
For $Cov(X,Y)$ you will need the $EXY$ from above.
To use CLT, your sample is large enough $n=35, V_n = \sum_{k=1}^{n}V_k, \mu=EV, \sigma = \sqrt{VarV}$:
$$
P\big(\frac{V_n}{n}>10) = P\bigg(\frac{\frac{V_n}{n}-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{10-\mu}{\frac{\sigma}{\sqrt{n}}}\bigg) \approx 1- \Phi\bigg(\frac{10-\mu}{\frac{\sigma}{\sqrt{n}}}\bigg)
$$
