The original question is:
A boy waits at the bus stop for a bus. The route that the boy wants to take offers buses that arrive in a Poisson process averaging $\lambda=0.5$ buses per hour. However, the boy has horrible luck and is easily distracted, making it is a guarantee that he will miss the first bus that arrives and forcing him to take the second. What is the likelihood the boy will have to wait over 3 hours in order to catch a ride?
I understand the Poisson distribution in general, but the caveat where he has to miss the first bus gets me confused.
Since $\lambda$ is non-integer given as buses per hour, I've tried to convert it into hours per bus, so that $\lambda=2$. So now one bus comes every two hours(?) But if he misses the first bus, does that mean $\lambda=4$ since he has to wait for the second one?
Any help on finding the probability would be appreciated.
Hint: You want to find the probability that the second arrival occurs after $3$ hours. This is equivalent to having either $0$ or $1$ arrival in the time interval $[0,3]$. The number of arrivals in $[0,3]$ follows the $\operatorname{Poisson}(3\lambda)$ distribution.