joint probability of random variables and order statistics I have a statistic question which I can intuitively guess the answer but I cannot find a good way to prove it. 
Suppose $N$ IID random variables $X_1, ..., X_N$ and their order statistics $X_{(1)}\leq ...\leq X_{(N)}$. Consider the probability
$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k}, X_1>\alpha_k)$
where $\alpha_1\leq \alpha_2\leq ... \leq \alpha_k$ are constants and $k\leq N$ is an integer.
I know how to compute the probability of the order statistic, so the only annoying part is $X_1$. Using the conditional probability, we can rewrite the above probability as
$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$
$=P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})*P(X_1>\alpha_k|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$
That is the place where I am stuck. From the conditional probability, we observe that $X_1$ has to be at least $(k+1)$th largest value. since the random variables are IID and the information of the order statistics do not make any $X$ "special", so $X_1$ has $\frac{N-k}{N}$ chance to meet the requirement. Therefore, purely based on intuition, The probability can be rewritten as
$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k}, X_1>\alpha_k)$
$=\frac{N-k}{N}P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$
I do not know if my guess is correct or not, and I would like to see a formal proof of the question. I will appreciate it if anyone can shed light on it. 
======================Update======================
A simulation shows that the conditional probability
$P(X_1>\alpha_k|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$
is not equal to 
$P(X_1$ is at least $(k+1)$th largest value$|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$
where the latter one is equal to $\frac{N-k}{N}$, so my guess is not correct. I'll appreciate anyone's help.
 A: NOT AN ANSWER
I just want to better specify the reasoning the brought me to the expression in the original comment, and correct some mistakes.
Let us start, for example, with $k=2$. We have $N$ independent identically distributed random variables $X_i$ with CDF $F(x)$, and $\alpha_1 < \alpha_2 $. If $X_{(i)}$ is the $i$th random variable after ordering, then we aim at computing
$$\mathcal P_{N,2}(\boldsymbol{\alpha})=P\left(X_{(1)} < \alpha_1 \land X_{(2)} < \alpha_2 \land X_1 > \alpha_2 \right).$$
Using conditional probabilities we have
$$\mathcal P_{N,2}(\boldsymbol{\alpha}) = P\left(X_1 > \alpha_2\right)P\left(X_{(1)} < \alpha_1 \land X_{(2)} < \alpha_2  | X_1 > \alpha_2 \right).\tag{1}\label{1}$$
Conditioning on $X_1$ implies that neither $X_{(1)}$ nor $X_{(2)}$ correspond to $X_1$. All the other variables, independently of the value assumed by $X_1$ can assume first and second position in the ordering. Therefore we can expand the second term in RHS of \eqref{1} by noting that we can select any of the $i=2,3,\dots,N$ variables as $X_{(1)}$ (it must assume a value smaller than $\alpha_1$) and then any of the other $N-2$ variables as $X_{(2)}$ (it must assume a value between $\alpha_1$ and $\alpha_2$). All the remaining $N-3$ variables need to have value greater than $\alpha_2$. We get
\begin{eqnarray}
\mathcal P_{N,2}(\boldsymbol{\alpha}) &=& \left[1-F(\alpha_2)\right] \cdot (N-1)(N-2)\cdot F(\alpha_1) \left[F(\alpha_2)-F(\alpha_1)\right]\cdot\left[1-F(\alpha_2)\right]^{N-3}=\\
&=&(N-1)(N-2) F(\alpha_1) \left[F(\alpha_2)-F(\alpha_1)\right]\left[1-F(\alpha_2)\right]^{N-2}.
\end{eqnarray} 
If this is correct, then we can extend the reasoning to any $k$:
$$\mathcal P_{N,k}(\boldsymbol{\alpha})=\frac{(N-1)!}{(N-k-1)!}F(\alpha_1)\left[1-F(\alpha_k)\right]^{N-k} \prod_{i=2}^{k}\left[F(\alpha_i)-F(\alpha_{i-1})\right]$$
