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