# Poisson process question about chance of 'points' meeting

I am learning about Poisson process and have this example.

Question:

Let there be a chip shop that opens at 5pm and thereafter customers arrive according to a Poisson process of rate 12 per hour.

(a) What is the probability that the first customer at the shop arrives between 5.15pm and 5.30pm? (b) Suppose each customer stays in the shop for 5 minutes. What is the probability that the second customer to arrive meets another customer at some point while visiting the shop?

My attempt

For the first part I found it awkward that they used hours starting at 5pm, so I said let the units for the Poisson process $$N=(N_t, t\geq 0)$$ be in minutes with $$t=0$$ referring to 5pm. Is this a good way to start? How would you do it?

Then I said the arrival time of the first customer, $$T_1$$ say, is exponential with rate 1/5 per min, so

$$\mathbb{P}($$first customer arrives between 5.15pm and 5.30pm$$) = \mathbb{P}(15

which I got after using the formula for the cdf of Exp, as $$e^{-3}-e^{-6}$$.

For the second part, I started by recalling the interarrival times $$Y_k=T_k-T_{k-1}$$ are iid exponential with rate 1/5 per min so,

$$\mathbb{P}($$2nd customer meets someone$$) = \mathbb{P}(Y_2<5$$ or $$Y_3<5)$$

which I got as $$1-\frac{1}{e^2}$$.

I am not confident as I am just starting out. Could anyone come up with a better reasoned answer? Better way to frame the question, ie better set up? Are these answers even right? As I am unsure. Thank you

These are the same results I found, by computing for part (a) \begin{align} \mathbb P(N(15)=0,N(30)-N(15)\geqslant 1) &= \mathbb P(N(15)=0)\mathbb P(N(30-N(15)\geqslant 1) \\ &= 1-e^{-\frac15\cdot15}\left(1-e^{-\frac15\cdot15}\right)\\ &= e^{-3}(1-e^{-3})\\ &= e^{-3}-e^{-6}, \end{align} and for part (b) (assuming without loss of generality, by the stationary increments property, that the first customer arrives at time $$0$$)
\begin{align} \small\mathbb P(N(5)>0) + \mathbb P(N(5)=0,N(10)-N(5)>0) &= \small1-\mathbb P(N(5)=0) + \mathbb P(N(5)=0)(1-\mathbb P(N(1)-N(5)=0) \\ &=\small 1-e^{-\frac15\cdot 5} +e^{-\frac15\cdot 5}\left(1-e^{-\frac15\cdot 5} \right) \\ &= 1-e^{-1}+e^{-1}(1-e^{-1})\\ &= 1-e^{-2}. \end{align}