The Statement of the Problem:
Consider two independent Poisson processes $X = \{X(t), t \ge 0\}$ and $Y = \{Y(t), t \ge 0\}$ where $E[X(t)] = \lambda t$ and $E[Y(t)] = \mu t$. Let two successive events of the $X$ process occur at $T$ and $T' > T$ so that $X(t) = X(T)$ for $T \le t < T'$ and $X(T') = X(T) + 1$. Define $N = Y(T') - Y(T)$ to be the number of events of the $Y$ process in the time interval $(T, T')$. Find the distribution of $N$.
Towards a Solution:
We are told that $$ \mathbb{P}\{N = m\} = \frac{\lambda}{\lambda + \mu}\left( \frac{\mu}{\lambda + \mu} \right)^m, $$ for $m= 0,1,2, ...$. I take it that the reader is supposed to derive this result.
Well, from the hint, it looks like $N \sim$ Geom$\left(\frac{\lambda}{\lambda + \mu}\right)$, where $m$ is the number of "failures" until the first "success" (and where a success occurs with probability $\frac{\lambda}{\lambda + \mu}$).
So, my reasoning was as follows. Let $T' - T = s$. Then, since $X$ and $Y$ are Poisson, they have independent and stationary increments; and so $$ \mathbb{P}\{N = m\} = \mathbb{P}\{Y(s) = m\} = \mathbb{P}\{Y(s) = m\}\mathbb{P}\{X(s) = 1\} , $$ since we know that $\mathbb{P}\{X(s) = 1\} = 1$. Now, I can just use the mass functions of $X(s)$ and $Y(s)$; i.e., $$ \mathbb{P}\{Y(s) = m\}\mathbb{P}\{X(s) = 1\} = e^{-\mu s}\frac{(\mu s)^m}{m!} e^{-\lambda s}\lambda s , $$ with the knowledge that $e^{-\lambda s}\lambda s = 1$. Of course, the equation $e^{-\lambda s}\lambda s = 1$ does not actually have any (real) solutions in $s$, so naturally I feel as if I'm on the wrong track here.
Given the form in which the solution is given, I assume I should be looking at waiting times for the processes, or something; but I'm not seeing how to incorporate those...
Think of it this way. Given that $T' - T = s$, what is the conditional distribution of $N$? That is to say, what is $$\Pr[N = m \mid T' - T = s]?$$ Well, this is easy: it must be Poisson with intensity $\mu s$, thus $$\Pr[N = m \mid T' - T = s] = e^{-\mu s} \frac{(\mu s)^m}{m!}, \quad m = 0, 1, \ldots.$$ All that remains is to compute the unconditional distribution for all $s > 0$: $$\Pr[N = m] = \int_{s=0}^\infty \Pr[N = m \mid T'-T = s]f_{T'-T}(s) \, ds,$$ where $f_{T'-T}(s)$ is the probability density for the interarrival time between successive Poisson events for process $X(t)$. But we know that this is exponentially distributed with intensity $\lambda$; i.e., $$f_{T'-T}(s) = \lambda e^{-\lambda s}, \quad s > 0.$$ So we simply integrate: $$\Pr[N = m] = \int_{s=0}^\infty e^{-\mu s} \frac{(\mu s)^m}{m!} \lambda e^{-\lambda s} \, ds = \lambda \mu^m (\lambda +\mu)^{-m - 1} = \frac{\lambda}{\lambda + \mu} \left( \frac{\mu}{\lambda + \mu} \right)^m.$$ This of course is the mechanical/computational approach. The heuristic approach is to consider without loss of generality that $T = 0$, and consider a "composite" Poisson process with total intensity $\lambda + \mu$, in which events are classified into one of two types, say types $X$ and $Y$ respectively. Type $X$ events occur with probability $\lambda/(\lambda + \mu)$, and type $Y$ events occur with probability $\mu/(\lambda + \mu)$. You can see that via Poisson thinning, these are in themselves Poisson processes with the desired respective intensities. Now, we want to model the number of type $Y$ events that occur before the first occurrence of a type $X$ event. This of course is a geometric random variable: $$\Pr[N = m] = (1-p)p^m, \quad m = 0, 1, 2, \ldots,$$ where $m$ is the probability of a type $Y$ event. Moreover, this generalizes quite nicely to the case where we are interested in counting the number of type $Y$ events that occur before the $x^{\rm th}$ occurrence of a type $X$ event, which would be negative binomial.