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(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?

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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.

Your calculation of part (c) is correct.

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