Let $X_1, X_2,...,X_n \sim \textrm{Expo}(1)$ distribution with $n \geq 3$. How to compute $\mathrm{P}(X_1 + X_2 \leq rX_3)$? So let $X_1, X_2,...,X_n$, where $n \geq 3$ be a random sample from the Expo(1) distribution. 
How do I set up the computation for 
$\mathrm{P}(X_1 + X_2 \leq rX_3 | \sum_{i=1}^n X_i = t)$ where $r, t > 0$? 
Edit: I do not know how to use the notion that $X_1 + X_2$ is independent from $\frac{X_1}{X_1 + X_2}$ to proceed with the problem. 

So I think here is a way of solving the question analytically:
$\mathrm{P}(X_1 + X_2 \leq rX_3 | \sum_{i=1}^n X_i = t)\\
\rightarrow \mathrm{P}\left(X_1 + X_2 + X_3 \leq (1 + r)X_3 | \sum_{i=1}^n X_i = t\right)\\
\rightarrow \mathrm{P}\left(\frac{X_1 + X_2 + X_3}{X_3} \leq (1 + r) | \sum_{i=1}^n X_i = t\right)\\
\rightarrow \mathrm{P}\left(\frac{X_3}{X_1 + X_2 + X_3} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\$
The sum of exponentially distributed r.v.'s follows a Gamma distribution. Hence, we have that
$X_1 + X_2 \sim \textrm{Gamma}(2,1)\\
X_3 \sim \textrm{Gamma}(1,1)$
And so,
$\mathrm{P}\left(\frac{X_3}{X_1 + X_2 + X_3} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\
\rightarrow \mathrm{P}\left(\frac{\textrm{Gamma}(1,1)}{\textrm{Gamma}(2,1) + \textrm{Gamma}(1,1)} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\
\rightarrow \mathrm{P}\left(\textrm{Beta}(1,2) \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)$
We can drop the conditional by noting that
$\frac{X_3}{X_1 + X_2 + X_3} \perp \sum_{i=1}^n X_i$. 
$U + V \perp \frac{U}{U+V}$ if $U$ and $V$ are Gamma distributed, and we let $U = X_1 + X_2$ and $V = X_3$.
To compute
$\mathrm{P}\left(\textrm{Beta}(1,2) \geq \frac{1}{1 + r}\right)$
we let a random variable $Z \sim \textrm{Beta}(1,2)$
$\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{1}{\textrm{Beta}(1,2)}z^{1-1}(1-z)^{2-1}\textrm{d}z\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{\Gamma(3)}{\Gamma(1)\Gamma(2)}z^{1-1}(1-z)^{2-1}\textrm{d}z\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{2!}{0!1!}z^{1-1}(1-z)^{2-1}\textrm{d}z\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 2z^{0}(1-z)^{1}\textrm{d}z\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 2-2z\textrm{d}z\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = (2z - z^2)|_{\frac{1}{1+r}}^1\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = 2 - 1 - \frac{2}{1+r} + \frac{1}{(1+r)^2}\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{-1+r}{1+r} + \frac{1}{(1+r)^2}\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{r^2 - 1}{(1+r)^2} + \frac{1}{(1+r)^2}\\
\rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{r^2}{(1+r)^2}$  
