I have a Question that I do not understand at all. If anyone can help it will be very much appreciated.
In the early hours of Easter Friday, News24 reported that 300 vehicles per hour entered KwaZulu Natal for the 2021 Easter Holiday.
Calculate the probability that no vehicles enter KwaZulu Natal in a given minute?
If you assume that the number of vehicles per unit of time can be represented by $X$, a Poisson r.v. with parameter $\lambda$, $X$'s PMF (probability mass function) will defined by: $$\text{Pr}(X=x)=\dfrac{e^{-\lambda} \lambda^x}{x!},\ \ x\in\{0,1,2,\ldots \}. $$
In this distribution, $\lambda$ is equal to $E(X)$ and $V(X)$.
As the problem asks for the probability of zero vehicles in a given minute, it is convenient to express $\lambda$ in term of vehicles per minute. As there are 300 vehicles per hour, there are $300/60=5$ vehicles per minute. Therefore $\lambda=5$ in this case. Now find $$\text{Pr}(X=0)=\dfrac{e^{-\lambda} \lambda^0}{0!}=e^{-5}\approx 0.00674.$$
An interesting question is checking whether Poisson is a reasonable assumption for the distribution in this case. There are goodness of fit tests you can use to address this verification. If you have data on the number of vehicles per minute, a simple quick and dirty procedure is checking whether the sample mean is close to the sample variance, as they are theoretically equal in the Poisson distribution.