Probability calculation involving order statistics There are $n+1$ independent and identically distributed random variables with the same distribution as $D \sim \text{Exp}(\mu)$, denoted by $D, D_1, D_2, \ldots, D_n$.
Define event $E_1$ as "$D$ is less than the $(\lfloor n/2 \rfloor + 1)$-order statistic of $D_1, D_2, \ldots, D_n$".
Define event $E_2$ as "$D$ is less than a specific (arbitrary but fixed) $D_k$ which is one of the first $(\lfloor n/2 \rfloor + 1)$ smallest of $D_1, D_2, \ldots, D_n$".

What are $\text{Pr}(E_2)$ and $\text{Pr}(E_2 \mid E_1)$?


My idea: The probability of $E_1$ is not hard to compute and if $n$ is given, I can calculate it.
However, I am quite confused about $\text{Pr}(E_2 \mid E_1)$. Is it just $\text{Pr}(D < D_k)$?
I am not sure about this, because I don't know whether the event $D < D_k$ is independent from $E_1$ or not. Probably not. I think $E_1$ contains some information about $D < D_k$. But what exactly is it?
 A: A short Mathematica program shows that $\text{Pr}(E_2) \neq \text{Pr}(D < D_k)$.
However, I still don't know how to calculate $\text{Pr}(E_2)$ analytically.
Taking $n=5$ and $\mu = 20$, we have 


*

*$\text{Pr}(E_1) = \text{Pr} \left(\text{Exp}(20) < (\lfloor n/2 \rfloor + 1) \text{-th order stat. of } \text{Exp}_{i}(20) \right) = \frac{1}{2}$

*$\text{Pr}(E_2) = \text{Pr} ( \text{Exp}(20) < \text{Exp}_{k}(20) < (\lfloor n/2 \rfloor + 1) \text{-th order stat. of } \text{Exp}_{i}(20)) = \frac{1}{7}$

*$\text{Pr}(D < D_k) = \text{Pr} ( \text{Exp}(20) < \text{Exp}_{k}(20) ) = \frac{1}{2}$

*$\text{Pr}(D < D_k \mid D_k < (\lfloor n/2 \rfloor + 1) \text{-th order stat. of } D_i) = \text{Pr} ( \text{Exp}(20) < \text{Exp}_{k}(20) \mid  \text{Exp}_{k}(20) < (\lfloor n/2 \rfloor + 1) \text{-th order stat. of } \text{Exp}_{i}(20)) = \frac{2}{7} $



Mathematica code for the fourth probability calculation:
Probability[
 x < y \[Conditioned] y < z,
 {
  Distributed[x, ExponentialDistribution[20]],
  Distributed[y, ExponentialDistribution[20]],
  Distributed[z, OrderDistribution[{ExponentialDistribution[20], 5}, {Floor[5/2] + 1}]]
  }
 ]

