Let $X_1, X_2, \dots, X_n$ be $n$ independent and identical random variables following the exponential distribution with mean $m$.
Let $h > 0$ be a fixed number and let $N \in \{1,2,\dots,n\}$ be the number of $\{X_i\}_{i=1}^n$ such that $X_i \leq d$. I was wondering if we can have a nice expression about the distribution of $N$, say, $\mathbb{P}[{N=i}]$ for $i \in \{0,1,2,\dots,n\}$?
My idea is to consider the order statistics of $\{X_i\}_{i=1}^n$. Let $Y_{(i)}$ be the $i$-th smallest number among $\{X_i\}_{i=1}^n$. Then $\mathbb{P}[{N=i}] = \mathbb{P}\{Y_{(i)} \leq d \mathrm{\;and\:}Y_{(i+1)}> d\}$.
Let $\{Z_i\}_{i=1}^n$ be $n$ independent random variables such that $Z_i$ follows the exponential distribution with mean $\frac{m}{n+1-i}$. Then we know $Y_{(i)} =^d \sum_{j=n-i+1}^n Z_i$. Thus $\mathbb{P}[{N=i}] = \mathbb{P}\{\sum_{j=n-i+1}^n Z_i \leq d \mathrm{\;and\:}\sum_{j=n-i}^n Z_i> d\}$.
I'm currently at this step. I'd like to clean the above expression further but I got stuck. I don't know if my idea is right not not or some other smarter way to calculate the distribution of $N$?
It's possible to find the distribution of $N$ without introducing the random variables $Z_i$: you are in essence performing $n$ independent trials, with the $i$th trial being a success if $X_i\leq d$. So $N$ is a binomial random variable with parameters $n$ and $p=1-e^{-\frac{d}{m}}$.