For clarity, this is the original problem statement in its entirety:
People walk into the post office at a rate of 5 people/hour. We consider the following two ways of modeling the number of people who arrive in a given hour:
Model A: We divide the hour into 50 time intervals of 72 seconds and assume at most 1 person arrives in each interval, and that arrival events are independent.
Model B: With the sample space Ω = Z≥0 = {0, 1, 2, . . . , } and probability measure given by the Poisson distribution with appropriate rate parameter λ.
For Model A, specify a sample space Ω and give a formula for the probability mass function.Then: Compute the probabilities of the following events using both Model A and Model B. Round your answers to 3 decimal places.
That there are no arrivals in the hour.
That there are at least two arrivals over the course of the hour.
I am very confused on where to even begin with this question, a walkthrough would be appreciated.
I will try to give an answer that better leads you to be able to solve this without giving away the final results. Let $\mathsf{N}$ denote the number of customers in the given hour.
In model A, there are $50$ independent times where a customer might arrive. Say that one customer arrives with probability $p$. Thus, the number of customers in each of the $50$ intervals can be modelled using a Bernoulli-distribution with parameter $p$ (it is either $0$ or $1$ by assumption), and the total number of customers in the five hours is the sum of these random variables, i.e. $\mathsf{N}=\sum_{i=1}^{50}\mathsf{X}_i$, where $\mathsf{X}_i\stackrel{\mathrm{i.i.d.}}{\sim}\mathrm{Bernoulli}(p)$. Do you recall what the distribution of a sum of independent Bernoulli-variables is? To calculate the probabilities, you must know the parameter $p$. We know that the rate of customers on average (I assume) is $5$, i.e. $\mathbb{E}[\mathsf{N}]=5$. Once you know the distribution of $\mathsf{N}$ here, this leads to a very simply equation in $p$ that can readily be solved.
In model B, we have $\mathsf{N}\sim\mathrm{Poisson}(\lambda)$ for some appropriate $\lambda$. Again, using that $\mathbb{E}[\mathsf{N}]=5$, finding $\lambda$ is not too difficult.
To calculate the final probability, simply remember that $$ \mathbb{P}(\mathsf{N}\ge2) =1-\mathbb{P}(\mathsf{N}<2) =1-\big(\mathbb{P}(\mathsf{N}=0)+\mathbb{P}(\mathsf{N}=1)\big) $$
Finally, if you want to check your answer I get the probability of the first event to be $6.74\cdot10^{-3}$ under model A and $5.14\cdot10^{-3}$ under model B.