This is a question I was working on from Phil Gregory's Bayesian Logical Data Analysis and I ended up getting stuck here.
The question seems like a simple Poisson rate-problem by taking $rate = 5/hour$ and then easily from that you can calculate $p(n=3|r,t)$ however the next question then specifies on average 5 photons are detected per hour which made me doubt my method for this question.
I then assumed that we didn't know the rate exactly and so this is my new attempt using Bayesian methods however it seems very long and I'm worried I've over-complicated it and actually the answer is very simple - please help!
$$ n_1 = 5 ; n_2 = 3 ; t_1 = t_2 = t = 1$$ $$ I = prior $$ $$p(n_2|n_1,I) = \int p(n_2|n_1,I,r) p(r|n_1,I) dr\tag{1}$$ where I am trying to marginalise over $r$.
$$p(r|n_1,I) = \frac{p(r|I)p(n_1|r,I)}{p(n_1|I)}\tag{2}$$ (Bayes' theorem) and so $$p(n_1|I) = \int^{1000}_0 p(r|I) p(n_1|r,I) dr \tag{3}$$ where I'm assuming a Jeffrey's prior for $r$ between 0 and 1000, so that $p(r|I) = \frac{k}{r}$ where $k$ is a constant. As it's a Poisson process, $p(n_1|r,I) = e^{-rt} \frac{rt^{n_1}}{n_1!} $
Doing this integral and subbing it back into eqn (2) produces (where I've set all t = 1) $$p(r|n_1,I) = e^{-r} \frac{r^{n_1-1}}{0.2 * n_1!}\tag{4}$$
Subbing this back into eqn (1) then gives $$p(n_2|n_1,I) = \int^{1000}_0 e^{-2r} \frac{r^{n_2+n_1-1}}{0.2 *n_2! *n_1!}dr $$ $$ \approx 0.137$$
This value seems a little low and so am I going about it in the right way or am I completely barking up the wrong tree?