# Clarification on Poisson distribution

(Poisson processes and queues) Consider a situation involving a server, e.g., a cashier at a fast-food restaurant, an automatic bank teller machine, a telephone ex- change, etc. Units typically arrive for service in a random fashion and form a queue when the server is busy. It is often the case that the number of arrivals at the server, for some specific unit of time $\lambda t$ can be modeled by a Poisson$(\lambda t)$ distribution and is such that the number of arrivals in nonoverlapping periods are independent. Suppose telephone calls arrive at a help line at the rate of two per minute. A Poisson process provides a good model.

(a) What is the probability that five calls arrive in the next 2 minutes?

(b) What is the probability that five calls arrive in the next 2 minutes and then five more calls arrive in the following 2 minutes?

(c) What is the probability that no calls will arrive during a 10-minute period?

Okay so, for the a) part we consider $\lambda=2, t=2, y=$ so we will get the Poisson distribution $\frac{(\lambda t)^y}{y}e^{-\lambda t} = \frac{(2 *2)^5}{5!}e^{-(2* 2)}$

For b) we will have to sum part a) with the new distribution of $\lambda = 2, t=4, y=5$

For part c) we will have another distribution of $\frac{{\lambda t}^0}{0!}e^{-(2*10)}$

Is this correct? Or for each part a), b) and c) we have to consider the new rate?

Your calculation of (a) is correct, but you should be more precise with your notation. Let $Y_t$ be the random number of calls that arrive in the next $t$ minutes. If $Y_t$ is a Poisson process with intensity/rate $\lambda$ per minute, then $$\Pr[Y_2 = 5] = e^{-\lambda t} \frac{(\lambda t)^5}{5!} = e^{-4} \frac{4^5}{5!}.$$
Your calculation of (b) is incorrect. You cannot add the intervals together to get $t = 4$, because when you do, you will no longer account for the fact that exactly $5$ events must occur within each $2$-minute interval. Instead, you must use the respective probability $\Pr[Y_2 = 5]$ and due to the independent increments property, you simply square this value to get the desired probability. In other words, if the probability that exactly $5$ events are observed in $2$ minutes is $p$, then the probability that you observe $5$ events between times $t = 0$ and $t = 2$ and another $5$ events between $t = 2$ and $t = 4$ is simply $p^2$. It would be wrong to compute $\Pr[Y_4 = 10]$, since this probability includes events where, for example, $3$ events occur in the first half and $7$ in the second half of the $4$-minute interval.