conditional probability exponential distributed 
Let $X_1,\ldots,X_n$ be independent random variables, $X_i \sim{}$
  $\mathrm{exponential}(\lambda_i)$. Let $X=\min\limits_{1\le i \le n} X_i$. Calculate
  $\mathbb{P}(X=X_i)$

At first I determined that $X\sim\mathrm{exponential}(\lambda_1+\cdots+\lambda_n)$. 
My next idea was that
$$\mathbb{P}(X=X_i\mid X\le x)=\frac{\mathbb{P}(X=X_i, X\le x)}{\mathbb{P}(X\le x)}=\frac{\mathbb{P}(X_i\le x)}{\mathbb{P}(X\le x)}=\frac{1-e^{\lambda_ix}}{1-e^{(\lambda_1+\cdots+\lambda_n)x}}$$
But I looked up in the solution and it should be  $\mathbb{P}(X=X_i)=\dfrac{\lambda_i}{\lambda_1+\cdots+\lambda_n}$ a.s.
Can someone help?
Thanks,
Zitrone
 A: We know the following about exponentials: PDF: $f_{X_i}(x) = \lambda_i e^{-\lambda_i x}$ and $P(X_i > x) = e^{-\lambda_i x}$. We will make use of them below. Consider 
\begin{split}
P(X_i  = X) 
&=
\int_{x=0}^{\infty} P(X_i = X, X \in (x-dx,x))
\\
&= 
\int_{x=0}^{\infty} P(X_i \in (x-dx, x), X_j > x\ \forall\ j\ne i)
\\
&=
\int_{x=0}^{\infty} P(X_i \in (x-dx, x)) \Pi_{j\ne i} P(X_j > x)
\\
&=
\int_{x=0}^{\infty} \lambda_i e^{-\lambda_i x} dx\ \Pi_{j\ne i} e^{-\lambda_j x}
\\
&=
\lambda_i \int_{x=0}^{\infty}  e^{- x \sum_{j=1}^n \lambda_j}dx
\\
&=
\frac{\lambda_i}{\sum_{j=1}^n \lambda_j}
\\
\end{split}
A: One can first clear up the situation, noting that, since the random variables $X_i$ are independent and absolutely continuous, there is almost surely no ex aequo, that is, $P(\exists i\ne j,X_i=X_j)=0$. Thus, the event $[X=X_i]$ is, up to some negligible events, equal to to the event $[\forall j\ne i,X_j\gt X_i]$.
The next step depends on the level of sophistication you are used to and, since you say nothing about this, it is difficult to continue but... 
...The most basic approach might be to consider the density $f_j$ of the distribution of each $X_j$ and to note that, by definition, $$p_i=P[\forall j\ne i,X_j\gt X_i]=\int_0^\infty f_i(x)\left(\prod_{j\ne i}\int_x^\infty f_j(x_j)\mathrm dx_j\right)\mathrm dx.$$  The value of the $j$th inner integral is $e^{-\lambda_jx}$ hence, introducing the parameter $\lambda=\lambda_1+\cdots+\lambda_n$, one gets, as desired, $$p_i=\int_0^\infty f_i(x)\exp\left(-\sum_{j\ne i}\lambda_j x\right)\mathrm dx=\int_0^\infty \lambda_i\exp\left(-\lambda x\right)\mathrm dx=\frac{\lambda_i}{\lambda}.$$
