An expression for $P(N_s=k|T_n=t)$ where {$N_t ; t\geq0$} is a Poisson process with parameter $\lambda$ Problem

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*Let {$N_t ; t\geq0$} be a Poisson process with parameter $\lambda$

*Solve for $P(N_s=k|T_n=t)$ where $T_n$ is the arrival time of the $n$-th event and $s$ and $k$ are positive integers

My Approach
So there are two cases where I have

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*Case 1: $s>t$ and $k\ge$ $n$

*Case 2: $s<t$ and $k\leq$ $n-1$
In Case 1, $P(N_s=k|T_n=t)=P(N_s=k)=\frac{{(\lambda}s)^k{e^{-{\lambda}s}}}{k!}$
In Case 2,  ($N_s|T_n=t$) ~ $B(n-1,s/t)$ so $P(N_s=k|T_n=t)$ = $\frac{(n-1)!}{k!(n-1-k)!}$$(\frac{s}{t})^k$$(1-\frac{s}{t})^{n-1-k}$
I am unsure in either case if my approach is correct and I also do not now how to prove that in Case 2, $(N_s|T_n=t)$ ~ $B(n-1,s/t)$
 A: Your result in the first case is not quite right.
The intuitive idea is that at time $t$, the first $n$ events have occured, and we are asking the probability that the remaining $k-n$ events happen in the remaining time $s-t$.
Since what happens in the remaining time is independent of what has happened until then, it should be the same as asking what the probability is of $k-n$ events happening in the first time $s-t$.
More stringently, we may write
$$
 \mathbb{P}(\mathsf{N}_s=k\mid \mathsf{T}_n=t)
 =\mathbb{P}(\mathsf{N}_t=n, \mathsf{N}_s-\mathsf{N}_t=k-n\mid \mathsf{T}_n=t)
 =\mathbb{P}(\mathsf{N}_{s-t}=k-n)
 =\frac{(\lambda(s-t))^{k-n}e^{-\lambda(s-t)}}{(k-n)!}.
$$
In the third step, several things happen.
Firstly, $\mathsf{N}_s-\mathsf{N}_t$ is independent of the process at time $t$, and we may split the probability into a product (since also the event $\{\mathsf{T}_n=t\}$ is dependent on the process only up until time $t$).
Secondly, the event $\{\mathsf{N}_t=n\}\cap\{\mathsf{T}_n=t\}$ is a sure event given the latter, so its conditional probability is $1$.
Finally, since again $\mathsf{N}_s-\mathsf{N}_t$ is independent of the process at time $t$, and hence of $\{\mathsf{T}_n=t\}$, the probability is the same whether we condition or not, and since $\mathsf{N}_s- \mathsf{N}_t\sim \mathsf{N}_{s-t}$ by stationary increments, the equality follows.
As for the second case, your result is correct.
Intuitively, given that at time $t$ there are $n$ events, where those events fall should not be given preference of one time over another, i.e. they should be uniform.
If this is the case, then each of the $n-1$ event times can be seen as a uniform random variable on the interval $[0, t]$, and whether they fall on one side of $s$ or the other can be viewed as independent Bernoulli variables with probability $s/t$.
To see that this is indeed the case, remember that the interarrival times of the process are exponentially distributed with parameter $\lambda$, i.e. $\mathsf{T}_k=\mathsf{X}_1+\mathsf{X}_2+\ldots+\mathsf{X}_k$ where $\mathsf{X}_i\overset{\mathrm{i.i.d.}}{\sim}\mathrm{Exponential}(\lambda)$.
As such, the joint density of the first $n$ arrival times by the chain rule is
\begin{align*}
 f_{\mathsf{T}_1,\ldots,\mathsf{T}_n}(t_1,\ldots,t_n)
 &=f_{\mathsf{T}_1}(t_1)f_{\mathsf{T}_2\mid \mathsf{T}_1}(t_2\mid t_1)\cdots f_{\mathsf{T}_n\mid \mathsf{T}_{n-1}}(t_n\mid t_{n-1}) \\
 &=(\lambda \mathrm{e}^{-\lambda t_1})(\lambda \mathrm{e}^{-\lambda(t_2-t_1)})\cdots(\lambda \mathrm{e}^{-\lambda(t_n-t_{n-1})}) \\
 &=\lambda^n \mathrm{e}^{-\lambda t_n},
\end{align*}
given that $t_1<t_2<\ldots<t_n$.
Also, since $\mathsf{T}_n\sim\Gamma(n, \lambda)$ (as the sum of i.i.d. exponentials), we find that the conditional distribution of the first $n-1$ event times given the $n$'th is
$$
 f_{\mathsf{T}_1,\ldots,\mathsf{T}_{n-1}\mid \mathsf{T}_n}(t_1,\ldots,t_{n-1}\mid t)
 =\frac{\lambda^n \mathrm{e}^{-\lambda t}}{\frac{\lambda^n}{(n-1)!}t^{n-1}\mathrm{e}^{-\lambda t}}
 =\frac{(n-1)!}{t^{n-1}}.
$$
This is exactly the order statistic of $n-1$ independent uniform random variables on the interval $[0, t]$, i.e. $\mathsf{T}_1,\ldots,\mathsf{T}_{n-1}\mid \mathsf{T}_n\sim \mathsf{U}^{(1)},\mathsf{U}^{(2)},\ldots,\mathsf{U}^{(n-1)}$, where $\mathsf{U}_1,\ldots, \mathsf{U}_{n-1}\overset{\mathrm{i.i.d.}}{\sim}\mathrm{Uniform}[0, \mathsf{T}_n]$ meaning that if we do not care about order (which we don't), we may as well view the first $n-1$ event times as being uniform on $[0, \mathsf{T}_n]$ as desired.
