How to prove $P\left\{\min(X_1,X_2,\dots,X_n) = X_i\right\} = \frac{\lambda _i}{\lambda_1+\dots+\lambda_n}$ , when $X_i$ is exponentially distributed I want to prove :
$P\left\{\min(X_1,X_2,\dots,X_n) = X_i\right\} = \frac{\lambda_i}{\lambda_1 + \dots + \lambda_n}$  when $X_i$ is exponentially distributed with parameter $\lambda_i$
I've made some progress by showing below:
$P(X_i > t) = P(X_1 > t)P(x_2>t)\dots P(X_n > t) = e^{-\lambda_1t}e^{-\lambda_2t}\dots e^{-\lambda_nt} = e^{-(\lambda_1 + \lambda_2 +\dots+ \lambda_n)t}$
but I don't know how to find the exact answer.
I'm sure the for final answer is should use Laplace transform am i right?
 A: Assuming the $\ X_j\ $ are independent,
\begin{eqnarray}
P\left(\,\min\left(X_1,X_2,\dots,X_n\,\right)=X_i\right)&=&P\left(X_i \le X_j\ \mbox{ for } j\ne i\right)\\
&=& \int_\limits{0}^\infty P\left(t\le X_j\ \mbox{ for } j\ne i\left|X_i=t\right.\right)\lambda_i e^{-\lambda_i t} dt\\
&=& \int_\limits{0}^\infty \prod_\limits{j\ne i}P\left(t\le X_j\right)\lambda_i e^{-\lambda_i t} dt\\
&=& \int_\limits{0}^\infty \lambda_i e^{-\sum_{j=1}^n\lambda_jt} dt\\
&=& \frac{\lambda_i}{\sum_\limits{j=1}^n\lambda_j}\ .
\end{eqnarray}
A: You forgot the assumption $X_1,X_2,\dots,X_n$ are independent.
So for $t>0$ and $\delta t$ small,
\begin{align*}
&\mathbb{P}(\min(X_1,\dots,X_n)=X_i\text{ and }X_i\in(t-\delta  t,t])\\
&=\mathbb{P}(X_i\in(t-\delta t,t])\prod_{j\neq i}[\mathbb{P}(X_j\geq t)+O(\delta t)]\\
&=[\lambda_i e^{-\lambda_i t}\delta t+o(\delta t)]e^{-\sum_{j\neq i}\lambda_j t}\\
&=\lambda_i e^{-(\lambda_1+\dots+\lambda_n)t}\delta t(1+o(1))
\end{align*}
Thus summing partition of $(0,\infty)$ into intervals of length $\delta t$, and taking $\delta t\to 0$,
$$
\mathbb{P}(\min(X_1,\dots,X_n)=X_i)=
\int_0^\infty\lambda_ie^{-(\lambda_1+\dots+\lambda_n)t}\,\mathrm{d}t=\frac{\lambda_i}{\lambda_1+\dots+\lambda_n}
$$
