If $x_1, x_2, \ldots , x_n \sim \operatorname{Exp}(\lambda)$, then find $P(x_1 \mid \sum_{i=1}^{n} x_{i}=t)$.
My approach is, $t=\sum_{i=1}^{n}x_i \sim \operatorname{Gamma}(n, \lambda)$, and $v=\sum_{i=2}^{n} x_{i} \sim \operatorname{Gamma}(n-1, \lambda)$. After that, I cannot proceed.
Let $X_1,\ldots,X_n$ be independent $\operatorname{Exp}(\lambda)$ variables. Also, let $T = \sum_{i=1}^{n} X_i$ and $V = \sum_{i=2}^{n} X_i$. Then the joint PDF of $(X_1, S)$ is computed as
$$ f_{X_1,T}(x_1, t) = f_{X_1}(x_1) f_{T\mid\{X_1=x\}}(t) = f_{X_1}(x_1)f_V(t-x_1), $$
and so,
\begin{align*} f_{X_1\mid\{T=t\}}(x_1) = \frac{f_{X_1}(x_1)f_V(t-x_1)}{f_{T}(t)} &= \frac{\lambda e^{-\lambda x_1} \cdot \frac{\lambda^{n-1}(t-x_1)^{n-2}e^{-\lambda(t-x_1)}}{(n-2)!}}{\frac{\lambda^n t^{n-1} e^{-\lambda t}}{(n-1)!}} \\ & (n-1) \frac{1}{t}\left(1-\frac{x_1}{t}\right)^{n-2}. \end{align*}
This means that $(X_1 \mid T = t)$ has the same distribution as $tU_{(1)}$, where $U_{(1)} \sim \operatorname{Beta}(1,n-1)$. In general, a similar computation shows that
$$ \left( \frac{X_1}{T}, \ldots, \frac{X_{n-1}}{T} \right) \stackrel{\text{d}}{=} (U_{(1)},\ldots,U_{(n-1)}),$$
where $(U_{(1)},\ldots,U_{(n-1)})$ are the order statistics of independent uniform variables $U_1, \ldots, U_{n-1}$ over the interval $(0, 1)$.