Stochastic process on truck arrival

Question

Trucks arrive at a particular toll booth on the highway according to Poisson process with rate one per minute. Assume midnight 12:00 AM as the start time of the process. Compute the probability that there are two arrivals during [01:01,01:04] AM. and three arrivals in the interval [01:03,01:05] AM

Since it's a Poisson process so X(t) follows Poisson( t ) distribution since rate is one given.

I was taking cases as number of arrivals that can happen in different time interval, e.g. two arrivals in [01:01,01:03] and three arrivals in [01:04,01:05] but it's getting very long.

Answer 1

Your approach is correct. Always try to divide the question to disjoint time segments. Here the segments are $I_1=[1:01,1:03]$, $I_2=[1:03,1:04]$ and $I_3=[1:04,10:05]$. Two events in $I_1\cup I_2$ and three events in $I_2\cup I_3$ can occur in the following cases $(I_1,I_2,I_3)$: $(0,2,1)$, $(1,1,2)$ or $(2,0,3)$. Can you continue from here?

Answer 2

It really is case-based. Sometimes that's just the way it is, but it's not really that bad since the cases can be split into $3$ groups such that in each group, the cases have the same probability.

Let $p(k)={\displaystyle{\frac{e^{-1}}{k!}}}$.

Note that $p(1)=p(0)$.

For $n\in \{1,2,3,4\}$, let $x_n$ be the number of trucks arriving in the $1$-minute interval starting at $n$ minutes after $01{\,:\,}00$.

Then we have the following cases . . . $$ \begin{array}{|c|c|c|c|c|c|} \hline \text{case}&x_1&x_2&x_3&x_4&\;\;\;\text{probability}\;\;\;\\ \hline 1&0&0&2&1& p(0)^2p(1)p(2)={\large{\frac{1}{2}}}e^{-4}\;\;\\ 2&0&1&1&2& p(0)p(1)^2p(2)=\text{same}\;\;\\ 3&1&0&1&2& p(0)p(1)^2p(2)=\text{same}\;\;\\ \hline 4&0&2&0&3& \;p(0)^2p(2)p(3)={\large{\frac{1}{12}}}e^{-4}\;\;\\ 5&2&0&0&3& p(0)^2p(2)p(3)=\text{same}\;\;\\ \hline 6&1&1&0&3& p(0)p(1)^2p(3)={\large{\frac{1}{6}}}e^{-4}\;\;\\ \hline \end{array} $$ Note that cases $1,2,3$ have the same probability, and similarly, cases $4,5$ have the same probability, hence the desired probability is $$ 3{\,\cdot} \Bigl( \frac{1}{2}e^{-4} \Bigr) \;+\; 2{\,\cdot} \Bigl( \frac{1}{12}e^{-4} \Bigr) \;+\; \frac{1}{6}e^{-4} \;=\; \frac{11}{6}e^{-4} $$