If $X_{1}, X_{2},\ldots, X_{n}$ be i.i.d random variables, what is the probability $$\mathbb{P}(\min( X_{1}, X_{2},\ldots,X_{n})=X_{1})\,?$$
Because every $X_{i}$ is equally likely to be the smallest one,
$$\mathbb{P}(\min( X_{1}, X_{2},\ldots,X_{n})=X_{1})=\frac1n$$
And my question is, if $X_{1}, X_{2},\ldots, X_{n}$ be independent random variables.
For example, the distribution of $X_{i}$ is exponential distribution with parameter $\lambda_{i}>0$. What is the probability $\mathbb{P}(\min( X_{1}, X_{2},\ldots,X_{n})=X_{1})$ ?
In the first part, the answer is $1/n$ if $X_i$ are continuous random variables. If they are discrete, the probability is greater than $1/n$ because there is nonzero probability that more than one $X_i$ is the minimum.
In the second part, the answer will obviously depend on the distributions of the $X_i$. If $X_i$ has density $f_i(x)$, $$ \mathbb P(\min(X_1,\ldots,X_n)=X_1) = \int_{-\infty}^\infty dx_1 \int_{x_1}^\infty dx_2 \ldots \int_{x_1}^\infty dx_n\; f(x_1)\ldots f(x_n) $$ In your exponential example (where $\lambda_i$ are rate parameters), this is $$ \int_0^\infty dx_1 \lambda_1 e^{-(\lambda_1+\ldots +\lambda_n) x_1}= \frac{\lambda_1}{\lambda_1+\ldots+\lambda_n}$$