I am looking to solve this problem. I am not sure if I did it correctly.
Let $N$ be a homogeneous Poisson process on $[0, \infty)$. Homogeneous, so it means it has linear mean value function. I am asked specifically to use the "order statistics property" to show, for $s < 1$ and $k \leq n$ $$ P(N(s) = k | N(1) = n ) = {{n}\choose{k}} s^k (1-s)^{n-k} $$
$\textbf{This is what I've come up with so far.}$ Note one has to use the relationship of the homogeneous Poisson process and the order statistics property.
Let $T_i$ be the arrival times, then $N(t) = \# \{ i \geq 1 : T_i \leq t \} = \sum_{i=1}^{N(t)} 1_{(T_i \leq t)}$
Then I can say, let $U_i \sim unif (0,1)$ $$ P(N(s) = k | N(1) = n ) = P(\sum_{i=1}^{N(s)} 1_{(T_i \leq s)} = k | N(1) = n) =P(\sum_{i=1}^{N(1)} 1_{(T_i \leq s)} = k | N(1) = n) \overset{\text{order stat.}}{=} P \left( \sum_{i=1}^{n} 1_{(U_i \leq s)} = k \right) $$ Since $1_{U_i \leq s} \sim Bern(s)$ then $Y_n = \sum_{i=1}^{n} 1_{(U_i \leq s)} \sim Bin(n,s)$, so $$ P \left( \sum_{i=1}^{n} 1_{(U_i \leq s)} = k \right) = {{n}\choose{k}} s^k(1-s)^{n-k} $$ $\textbf{Is this correct?}$