1. **Problem Statement:** The number of cars arriving at the station in an hour follows a Poisson distribution with an average rate $\lambda = 5$ cars per hour. 2. **Poisson Distribution Formula:** The probability of observing $k$ events in a fixed interval is given by: $$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda$ is the average rate and $k$ is the number of events. 3. **Part (a): Probability of exactly one car arriving ($k=1$):** $$P(X=1) = \frac{5^1 e^{-5}}{1!} = 5 e^{-5}$$ Calculate the value: $$P(X=1) = 5 \times e^{-5} \approx 5 \times 0.0067379 = 0.03369$$ 4. **Part (b): Probability of at least six cars arriving ($k \geq 6$):** This is the complement of having five or fewer cars: $$P(X \geq 6) = 1 - P(X \leq 5) = 1 - \sum_{k=0}^5 \frac{5^k e^{-5}}{k!}$$ Calculate the cumulative probability $P(X \leq 5)$: $$P(X \leq 5) = e^{-5} \left( \frac{5^0}{0!} + \frac{5^1}{1!} + \frac{5^2}{2!} + \frac{5^3}{3!} + \frac{5^4}{4!} + \frac{5^5}{5!} \right)$$ Calculate each term: $$= e^{-5} (1 + 5 + 12.5 + 20.8333 + 26.0417 + 26.0417) = e^{-5} \times 91.4167$$ $$P(X \leq 5) \approx 0.0067379 \times 91.4167 = 0.6160$$ Therefore, $$P(X \geq 6) = 1 - 0.6160 = 0.3840$$ **Final answers:** - (a) $P(X=1) \approx 0.0337$ - (b) $P(X \geq 6) \approx 0.3840$