Poisson Process and Conditional Probability Let $X= (X(t); t\ge0)$ be a poisson process with the intensity ($\lambda$ per hour)
A) Find the conditional probability of having $m$ events in the first $t$ hours, given that there were $n$ events in first $T$ hours. (here $0\le m \le n$ and $0<t<T$)
I need someone to check if this is right.
$P(m$ events in $t$ hours | $n$ events in $T$ hours) = $P(m$ events in $t$ hours)/$P(n$ events in $T$ hours) = 
$$ {((e^{-\lambda t})(\lambda t)^m) }/m! \over {((e^{-\lambda T})(\lambda T)^n) /n!} $$
Any help would be greatly appreciated.
 A: Hint: In order for you answer to be correct you also need to add the following term in the numerator $$\begin{align*}&\phantom{\,\=}P(m \text{ events in $t$ hours | $n$ events in $T$ hours})= \\ \\& =\frac{P(m \text{ events in $t$ hours, $n-m$ events in $T-t$ hours})}{P(n \text{ events in $T$ hours})}=\\\\& =\frac{P(m \text{ events in $t$ hours})\cdot P(n-m \text{ events in $T-t$ hours})}{P(n \text{ events in $T$ hours})}=\ldots\end{align*}$$ 
Then the formula you used for the poisson distribution is correct. The result should reflect the binomial distribution (see second bullet). 
A: Hint: Let $B$ be the event "$n$ events in the first $T$" and $A$ the event $m$ events in the first $t$."
We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$. 
To compute $\Pr(A\cap B)$, note that for this event to occur we need $m$ events in the first $t$ hours and $n-m$ additional  events in the remaining $T-t$ hours. Thus you need to multiply your current numerator by a suitable factor.
Then simplify. The answer will look quite familiar.
