A pedestrian wishes to cross a one-way street where cars pass by according to a Poisson process with rate $10$ per minute. The pedestrian will only cross when there are no cars for $10$ seconds. If the pedestrian arrives at the stop at a random time, what's the probability distribution for the number of cars that will pass before you can cross?
I did hear about memorylessness of Poisson Process, so the result should reflect that, but I'm new to Poisson processes so I cannot solve this one.
I know the number of cars that pass by has rate $10/60 = 1/6$ per second. So we want to find the expected time for $10$ arrivals. Let $N(t)$ be the number of arrivals that have occured by time $t$. Then I think we want $P(\min\{t' \mid N(t) = 10'\} \leq t')$.
But this looks really messy. Maybe there is a better way to do it?
Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda>0$ and fix $T>0$. Then $$ \mathbb P(N(T)=n) = e^{-\lambda T}\frac{(\lambda T)^n}{n!},\ n=0,1,2,\ldots. $$ Here $\lambda=\frac16$ and $T=10$ so we have $$ \mathbb P(N(10)=n) = e^{-\frac16\cdot10}\frac{\left(\frac16\cdot10\right)^n}{n!} = e^{-\frac53}\frac{\left(\frac53\right)^n}{n!},\ n=0,1,2,\ldots. $$