I am not sure I totally understand Poisson distributions..
6.2 Assume that the number of uninspected cars caught at a state police checkpoint is Poisson distributed with average 2.1 per hour. Find:
λ=average=2.1 cars/hour
a. P(no cars caught in a 1 hour stretch) $$=\sum^{0}_{j=0}\frac{2.1^j}{j!}e^{-2.1} =\frac{2.1^0}{0!}e^{-2.1})$$
b. P(at least 2 cars caught in 1 hour) = 1 - P(1 car caught in 1hour) \begin{align} & =1-\sum^1_{j=0}\frac{2.1^j}{j!}e^{-2.1} \\ & =1-(\frac{2.1^0}{0!}e^{-2.1}+\frac{2.1^1}{1!}e^{-2.1}) \\ & =1-(\frac{1}{1}e^{-2.1}+\frac{2.1}{1}e^{-2.1}) \\ & =1-(e^{-2.1}+2.1e^{-2.1}) & =1-3.1e^{-2.1} \\ & =1-3.1(0.1224) \end{align}
c. P(more than 5 caught in one hour) = 1 - P(5 cars caught in 1hour)
$$=1-\left(\frac{2.1^5}{5!}e^{-2.1}+\frac{2.1^4}{4!}e^{-2.1}+\frac{2.1^3}{3!}e^{-2.1}+\frac{2.1^2}{2!}e^{-2.1}+\frac{2.1^1}{1!}e^{-2.1}+\frac{2.1^0}{0!}e^{-2.1}\right)$$
Answers to problem are correct. However in order to understand Poisson rate λ which in your example it represents average cars coming per hour, you may try to ask the following two questions. what are average cars coming in half hour and in two hours? Since Poisson rate is linear and memoryless they are λ/2 and 2*λ.Hope this help you understand more about this Poisson rate.