I am having some issues finding a resource to assist me with this. We have a Poisson distribution with on average sees 5 trains roll through its station every 1 hour. Let the random variable X be the time in hours between arrivals of consecutive trains. Find the P(X = 1/3).
I suppose I could break it down into minutes instead, then at the end of the problem work my way back up to hours but otherwise I am unable to use the general formula for solving probability.
In a Poisson process, two different kinds of random variables are involved:
The probability that the time is exactly $1/3$ hour is $0$.
The probability that the time is at least $1/3$ hour is the same as the probability that the number of trains in that $1/3$ hour is $0$. The expected number of trains in $1/3$ hour is $5/3.$ The probability that the number of trains during that time is $0$ is therefore $\dfrac{(5/3)^0 e^{-5/3}}{0!} = e^{-5/3}.$