Suppose that radioactive particles hit a surface according to a Poisson process of rate $\lambda = 20$ per hour. A total of $100$ particles arrived in the first $10$ hours. Conditional on this, I find the probability that $n(n ≤ 100)$ arrived the first $4$ hours.
The answer is:
\begin{align} P(X(4)=n|X(10) = 100) &= \frac{P(X(4)=n,X(10) = 100)}{P(X(10) = 100)} \\ &= \frac{P(X(4)=n,X(10)-X(4) = 100-n)}{P(X(10) = 100)} \\ &= \frac{P(X(4)=n)\cdot P(X(10)-X(4) = 100-n)}{P(X(10) = 100)} \\ &= {100\choose n}\left(\frac{4}{6}\right)^n\left(\frac{6}{10}\right)^{100} \end{align}
Now, if plot this for values $n=0$ to $n=100$, I get the following graph:
Is there a reason for this? I know that the Poisson Distribution can be approximated by a normal distribution, but is there a reason this conditional process is as well?
As pointed out in the comments, the conditional distribution is not normal, but binomial with $n=100$ and $p=\frac 4{10}$. Maybe this is less surprising if you know the Poisson-limit theorem: https://en.wikipedia.org/wiki/Poisson_limit_theorem
For a moment, think of the Poisson distribution as a Binomial distribution: then each of the 100 particles (which arrived upon time 10) have an equal chance of arriving at any of the times $1,2,\dots,10$. Then the expression for the conditional probability is less surprising.
Hope that this was helpful.