Smallest order statistics Suppose $X_1$ has exponential distribution with mean $\frac{1}{\theta}$ and
$X_2,\ldots,X_n$ have exponential distribution with mean $\frac{2}{\theta}$. also suppose
$X_1,X_2,\ldots,X_n$ are independent.how likely that $X_1$ be smallest order statistics
 in sample $X_1,X_2,\ldots,X_n$? 
 A: Taking André Nicolas's comment, replace $X_1$ by the lower of independent $Y$ and $Z$ each with exponential distribution with mean $\frac{2}{\theta}$.  The minimum of $Y$ and $Z$ has an exponential distribution with mean $\frac{1}{\theta}$, the same distribution as $X_1$.
So now you are asking what is the probability that either $Y$ or $Z$ is the lowest of $Y,Z, X_2,\ldots, X_n$, where all are continuous iid. By symmetry this is $$\frac{2}{n+1}$$
A: $$
\Pr(x < X_2\  \&\  x<X_3\  \&\  \cdots\  \&\  x<X_n) = \Pr(x<X_2)\cdots\Pr(x<X_n) = \left( e^{-\theta x/2} \right)^{n-1}.
$$
Therefore
$$
\begin{align}
& {} \quad \Pr(X_1 < X_2\  \&\  X_1<X_3\  \&\  \cdots\  \&\  X_1<X_n) \\  \\
&  = \mathbb{E}(\Pr(X_1 < X_2\  \&\  X_1<X_3\  \&\  \cdots\  \&\  X_1<X_n \mid X_1)) = \mathbb{E}\left( \left(e^{-\theta X_1/2}\right)^{n-1} \right) \\  \\
& = \int_0^\infty \left( e^{-\theta x/2}\right)^{n-1} \cdot e^{-\theta x} \; \theta\;dx = \int_0^\infty e^{-(n+1)\theta x/2} \; \theta\;dx= \int_0^\infty e^{-(n+1)u} \; 2 \; du = \frac{2}{n+1}.
\end{align}
$$
(I've just revised this after Henry pointed out an obvious flaw that I had neglected.  I shall return shortly to check the details.)
