Finding the probability that at most n events take place on any interval during a given time period
Generating Function Approach
Let's compute the probability that no $5$ minute span has more than $1$ car entry.
Poisson with $\bf{1}$ Minute Resolution
Poisson says the probability that $0$ cars enter the street in a given minute is $e^{-\lambda}$, and the probability that $1$ car enters the street in a given minute is $\lambda e^{-\lambda}$.
Let a minute with one car be represented by '${+}$', which has a probability of $\lambda e^{-\lambda}$. Let a minute with no cars be represented by '${-}$', which has a probability of $e^{-\lambda}$.
The $60$ minute span can be uniquely built from atoms of '${-}$' and '${+}{-}{-}{-}{-}$' followed by an optional terminal atom of '$\text{+}$', '$\text{+}{-}$', '$\text{+}{-}{-}$', '$\text{+}{-}{-}{-}$'
This gives a generating function of $$ f_\lambda(x)=\frac{1+\overbrace{\ \ \lambda e^{-\lambda}x\ \ }^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}}{1-\underbrace{\ \ \ e^{-\lambda}x\ \ \ }_{-}-\underbrace{\lambda e^{-5\lambda}x^5}_{+----}}\tag1 $$ The probability that in no $5$ minute span of the hour there is more than $1$ car is $$ \begin{align} \left[x^{60}\right]f_\lambda(x) &=e^{-60\lambda}\left(1+60\lambda+1540\lambda^2+22100\lambda^3+194580\lambda^4\right.\\ &\left.+1086008\lambda^5+3838380\lambda^6+8347680\lambda^7+10518300\lambda^8\right.\\[2pt] &\left.+6906900\lambda^9+1961256\lambda^{10}+167960\lambda^{11}+1820\lambda^{12}\right) \end{align}\tag2 $$ Computing the complementary probability, we get the probability that two or more cars will enter in a single $5$ minute period during an hour:
The Generating Function
In order to prevent any $5$ minute span from having more than $1$ car entry, above, we break down the $60$ minutes into $1$ minute atoms:
The $60$ minute span can be uniquely built from atoms of '${-}$' and '${+}{-}{-}{-}{-}$' followed by an optional terminal atom of '$\text{+}$', '$\text{+}{-}$', '$\text{+}{-}{-}$', '$\text{+}{-}{-}{-}$'
and we compute the probabilities for each of these atoms using the Poisson Distribution:
Poisson says the probability that $0$ cars enter the street in a given minute is $e^{-\lambda}$, and the probability that $1$ car enters the street in a given minute is $\lambda e^{-\lambda}$.
In the generating function, we represent an '${+}$' by $\lambda e^{-\lambda}x$ and an '${-}$' by $e^{-\lambda}x$. Thus, the power of $x$ counts the minutes and the coefficients accumulate the probabilities. Thus, for each of these atoms, we include a factor of $$ \overbrace{\phantom{i^5}e^{-\lambda}x\phantom{i^5}}^{-}+\overbrace{\lambda e^{-5\lambda}x^5}^{+----}\tag3 $$
To account for any number of the standard atoms ('${+}$' and '${+}{-}{-}{-}{-}$'), we total the product of all powers of the standard atom: $$ \sum_{k=0}^\infty\left(e^{-\lambda}x+\lambda e^{-5\lambda}x^5\right)^k=\frac1{1-e^{-\lambda}x-\lambda e^{-5\lambda}x^5}\tag4 $$ Since products of the standard atoms of length $4$ or more will end in at least $4$ '${-}$'s, we add optional atoms of '${+}$', '${+}{-}$', '${+}{-}{-}$', or '${+}{-}{-}{-}$'. This is included in the generating function as a factor of $$ \overbrace{\phantom{iw^2}1\phantom{iw^2}}^\text{none}+\overbrace{\phantom{i^2}\lambda e^{-\lambda}x\phantom{i^2}}^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}\tag5 $$ Thus, we arrive at the generating function, $f_\lambda(x)$, given in $(1)$. To get the probability for a span of $60$ minutes, we look at the coefficient of $x^{60}$: $\left[x^{60}\right]f_\lambda(x)$.
Poisson with $\boldsymbol{\frac1n}$ Minute Resolution
We can handle finer resolutions using atoms of '${-}$' and '${+}\overbrace{{-}{-}\cdots{-}}^{5n-1}$'. To simplify notation, we will write $e^{-\lambda/n}x=u$. $$ \begin{align} g_{\lambda,n}(x) &=\frac{1+\frac\lambda{n}e^{-\lambda/n}x+\frac\lambda{n}e^{-2\lambda/n}x^2+\cdots+\frac\lambda{n}e^{-(5n-1)\lambda/n}x^{5n-1}}{1-e^{-\lambda/n}x-\frac\lambda{n}e^{-5\lambda}x^{5n}}\tag{6a}\\ &=\frac{1+\frac\lambda{n}u\,\frac{1-u^{5n-1}}{1-u}}{1-u-\frac\lambda{n}u^{5n}}\tag{6b}\\ &=\frac{\left(1-u-\frac\lambda{n}u^{5n}\right)+\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)\left(1-u\right)}\tag{6c}\\ &=\frac1{1-u}+\frac{\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)(1-u)}\tag{6d}\\ &=\frac1{1-u}+\left(\frac1{1-u-\frac\lambda{n}u^{5n}}-\frac1{1-u}\right)\frac1{u^{5n-1}}\tag{6e} \end{align} $$ Since all coefficients of $\frac1{1-u}$ are $1$, we get $$ \begin{align} \left[x^{60n}\right]g_{\lambda,n}(x) &=e^{-60\lambda}\!\left[u^{60n}\right]g_{\lambda,n}(x)\tag{7a}\\[9pt] &=e^{-60\lambda}\!\left[u^{65n-1}\right]\frac1{1-u-\frac\lambda{n}u^{5n}}\tag{7b}\\ &=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\left(u+\frac\lambda{n}u^{5n}\right)^k\tag{7c}\\ &=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}u^{k-j}\left(\frac\lambda{n}\right)^ju^{5nj}\tag{7d}\\ &=e^{-60\lambda}\sum_{j=0}^{12}\binom{65n-1-(5n-1)j}{j}\left(\frac\lambda{n}\right)^j\tag{7e} \end{align} $$ For $n=1$, $\text{(7e)}$ agrees with $(2)$.
Infinite Resolution
The limit as $n\to\infty$ of $\text{(7e)}$ is
$$
\begin{align}
p(\lambda)
&=e^{-60\lambda}\sum_{j=0}^{12}\frac{(65-5j)^j}{j!}\lambda^j\tag{8a}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{8b}
\end{align}
$$
At $\frac14$ car per minute, the probability is $99.91\%$ that two or more cars will enter in a single $5$ minute period during an hour.
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.
