# poisson process an probability

Calls to an emergency ambulance service in a large city are generated by a Poisson process with λ = 3 calls per hour.

How can I find the nature of the probability distribution for the lengths of time between successive calls to the ambulance service? and also how can I Obtain the mean length of time between successive calls to the ambulance service

I imagine that in your text/notes there is information about the relationship between a Poisson process and the exponential distribution. If you have a Poisson process with parameter (mean) $\lambda$, then the waiting time between two successive events has exponential distribution with parameter (this time, not mean) $\lambda$.
In fact the waiting time $T$ has density function $f_T(t)=\lambda e^{-\lambda t}$ for $t \ge 0$, and $f_T(t)=0$ for $t<0$.
The mean of a random variable with this exponential distribution is $1/\lambda$. In your situation, $\lambda=3$, so the mean waiting time is $1/3$.
If you want to find the probability that the waiting time is $\le w$, the answer is not hard to find. It is $$\int_0^w \lambda e^{-\lambda t} dt$$ The integration is not difficult. One antiderivative (indefinite integral) of $\lambda e^{-\lambda t}$ is $-e^{-\lambda t}$, so plugging in the end-points gives you that the probability that the waiting time between successive calls is $\le w$ is given by $$1 - e^{-\lambda w}$$
The relationship between the Poisson and the exponential is pretty fundamental. I am sure that it is discussed in your learning materials. Do remember the relationship between the means. For a Poisson, $\lambda$ large means lots of "calls" are happening per hour. So it makes sense that if $\lambda$ is large, the mean waiting time between calls is small. In fact this mean waiting time is $1/\lambda$.