Suppose that cars pass a certain street location according to a Poisson process with
rate λ and that it takes a person $10$ units of time to cross the street.
Suppose that when arriving at the street, the person looks at the coming cars and
waits until the first time when there is a gap of at least $10$ units of time between cars to
cross. What is the expected number of cars that the person has to wait to pass until
it is safe to cross.
It's a modification of the average waiting time which I was able to solve but not able to get an idea about how to get the expected number of cars.
Best Answer
Either the next car is at least $10$ units away or it is not. These gaps follow an exponential distribution with rate $\lambda$, and the probability a sample from said distribution is greater than $10$ is $e^{-10\lambda}$. Thus the number of cars the person has to let pass before crossing is a geometric distribution starting from $0$ (since they might already be able to cross) with success probability $p=e^{-10\lambda}$, and its expected value is $\frac1p-1=e^{10\lambda}-1$.