The probability density function (pdf) for a train arriving at any time from initial time $T_0$ is normally modelled by a Poisson distribution with rate $\lambda$. That is given $X$, a random variable representing time since $T_0$: $$X\sim Poisson(\lambda)$$
As a specific example, suppose a train is required to arrive between 7:00 AM and 7:05 AM. One and only one train is allowed to arrive in this interval. What is the pdf of such a train arriving at any given time $t \in [0,5]$ in minutes.
If a Poisson distribution is used, the rate would be $1/5 = 0.2$ per minute. However, the Poisson distribution is used in domain of $t\in[0,\infty)$, therefore leading to some trunication. Furthermore, since the area under the graph is not 1 anymore, this also requires 're-normalizing', which I am not aware of any mathematical proof that this is actually valid.
So, what is the probability of a train arriving in 5-minute interval given one and only one MUST arrive every 5-minute interval?
At first glance, this problem looks ill-posed, since nothing told you the mechanism for enforcing the rule that exactly one train must arrive in every 5-minute interval, combined with Poisson distributions of arrival. (For example, you could say that trains arrive at Poisson distributed times, except that after a train arrives there is a five minute gap and then the Poisson rate resumes, and except that if there is a 5-minute gap, a train is forced to arrive immediately.
However, in fact the problem is well-posed, and even trivial. If one and only one train must arrive in any 5-minute interval, then the trains must arrive exactly 5 minutes apart, and the probability of a train arriving in any given 5-minute interval is exactly 1.