How to prove this property in a Poisson process? For a Poisson process show, for $s < t$, that
$P(N(s)=k|N(t)=n) = \binom{n}{k} (\frac{s}{t})^k (1-\frac{s}{t})^{n-k}$
 A: The joint probability mass function for $(N(s), N(t))$ for $s<t$ is, using the property of independent increments, i.e. $N(t) \sim N(s) + N(t-s)$:
$$
  \mathbb{P}(N(s) = k, N(t) = n) = \mathbb{P}(N(s) = k, N(t) - N(s) = n - k) = \\\frac{(\mu s)^k}{k!} \mathrm{e}^{-\mu s} \cdot \frac{(\mu (t-s))^{n-k}}{(n-k)!} \mathrm{e}^{-\mu(t-s)} \cdot [ n-k \geqslant 0, k \geqslant 0] 
$$
Then, assuming $n \geqslant k \geqslant 0$:
$$
  \mathbb{P}(N(s) =k | N(t) = n) = \frac{\mathbb{P}(N(s)=k,N(t)=n)}{\mathbb{P}(N(t)=n)} = \\
     \frac{\frac{\mu^k s^k}{k!} \cdot \frac{\mu^{n-k} (t-s)^{n-k}}{(n-k)!} \cdot \mathrm{e}^{-\mu t} }{ \frac{\mu^n t^n}{n!} \mathrm{e}^{-\mu t}} = 
    \left( \frac{s}{t} \right)^k \left( 1- \frac{s}{t} \right)^{n-k} \binom{n}{k}
$$
A: Let 
$\ \ \ \ A$ be the number of events occurring in the interval $[0,s]$, 
$\ \ \ \ B$ be the number of events  occurring in the interval $[s,t]$, 
and 
$\ \ \ \ C$ be the number of events  occurring in the interval $[0,t]$.
Using the fact that in a Poisson process, events occurring in one time frame are independent of events occurring in a different time frame, we may write the sought after probability as
$$\tag{1}
\eqalign{P ( A  =k \mid C= n) &={P\bigl((A=k)\cap ( C=n) \bigr)\over P(C=n)}\cr
&={P\bigl((A=k)\cap ( B=n-k) \bigr)\over P(C=n)}\cr
&={P (A=k)   P ( B=n-k)  \over P(C=n)}
.}
$$
Now, if the Poisson process has parameter $\lambda$, then
$\ \ \ \ A$ has Poisson distribution with parameter $s\lambda$, 
$\ \ \  \ B$ has Poisson distribution with parameter $(t-s)\lambda$, 
and 
$\ \ \ \ C$ has Poisson distribution with parameter $t\lambda$.
So, using equation $(1)$:
$$\eqalign{
  P( A  =k \mid  C= n)  &={  {(s\lambda)^k e^{-s\lambda}\over k!}
{( (t-s)\lambda)^{n-k} e^{-(t-s)\lambda}\over (n-k)!} \over {(t\lambda)^n e^{-t\lambda}\over n!}}\cr
&={n\choose k} {s^k(t-s)^{n-k} \lambda^k \lambda^{n-k} e^{-t\lambda}\over t^n\lambda^ne^{-t\lambda}}\cr
&={n\choose k} {s^k(t-s)^{n-k}\over t^kt^{n-k}}\cr
&={n\choose k}  {\Bigl({s\over t}\Bigr)^k \Bigl(1-{s\over t}\Bigr)^{n-k}}.\cr
}
$$ 
