Finding the probability that at most n events take place on any interval during a given time period I am interested in the general case, but let us start with a smaller example. Suppose that cars arrive to a street with intensity $\lambda$ per minute. We would like to know the probability that at least two cars have arrived on the street during any five minute interval in the next hour. How can we find this?
My initial thought was to use complementary event: 1 - the probability that at most one car is on the street in any five minute period during the next hour. Hence my reasoning was something like $1 - \int_0^{55}\mathbb{P}(N(s + 5) - N(s) \leq 1)ds - \int_{55}^{60}\mathbb{P}(N(60) - N(55 + s) \leq 1)ds$.
But I quickly realized that finding this probability might not be that easy, since the intervals we are considering overlap, namely $N(5)$ and $N(5 + s)$ overlap except for the infinitesimal point $s$. So then, is the correct way to integrate just $\mathbb{P}(N(s) \leq 1$ over the region? Moreover, I think that my line of reasoning is missing something critical, since I do not see a reason, why the summation would not end up being negative.
 A: Conditional Probability Approach
Poisson says that in an hour, the probability that $n$ cars have entered is
$$
e^{-60\lambda}\frac{(60\lambda)^n}{n!}\tag1
$$
Given that $n$ cars have entered in the hour, the probability that none is within $5$ minutes of another is the ratio of the volumes of the simplices
$$
\frac{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60-5(n-1)\right\}\right|}{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60\right\}\right|}=\left(\frac{13-n}{12}\right)^n\tag2
$$
Each $x_k$ is the time from car $k-1$, or the beginning of the hour, to car $k$. With $n-1$ buffers of $5$ minutes removed, we get the numerator; without the buffers removed, we get the denominator.
Thus, Bayes' Theorem  gives
$$
\begin{align}
p(\lambda)
&=\sum_{n=0}^{12}e^{-60\lambda}\frac{(60\lambda)^n}{n!}\left(\frac{13-n}{12}\right)^n\tag{3a}\\
&=e^{-60\lambda}\sum_{n=0}^{12}\frac{(65-5n)^n}{n!}\lambda^n\tag{3b}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\[3pt]
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\[3pt]
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{3c}
\end{align}
$$
This matches the result from $(8)$ of my previous answer.

