1. **Problem statement:** We have an average rate of 150 cars per hour passing a point. We want to find probabilities related to the number of cars passing in different time intervals, assuming a Poisson process. 2. **Formula and explanation:** The number of cars passing in a time interval follows a Poisson distribution with parameter $\lambda = \text{rate} \times \text{time}$. The Poisson probability mass function is: $$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $k$ is the number of events (cars), and $\lambda$ is the expected number of events in the interval. 3. **Calculate the rate per minute:** Since 150 cars pass per hour, the rate per minute is: $$\lambda_{1\text{ min}} = \frac{150}{60} = 2.5$$ cars per minute. --- **i. Probability of congestion in any one minute (more than 6 cars):** We want $P(X > 6)$ where $X \sim \text{Poisson}(2.5)$. Calculate: $$P(X > 6) = 1 - P(X \leq 6) = 1 - \sum_{k=0}^6 \frac{2.5^k e^{-2.5}}{k!}$$ --- **ii. Probability that from 8 to 13 cars pass in any eight-minute period:** First, find $\lambda$ for 8 minutes: $$\lambda_{8\text{ min}} = 2.5 \times 8 = 20$$ We want: $$P(8 \leq X \leq 13) = \sum_{k=8}^{13} \frac{20^k e^{-20}}{k!}$$ --- **iii. Probability that at most 2 cars pass in any 30-second period:** 30 seconds is 0.5 minutes, so: $$\lambda_{0.5\text{ min}} = 2.5 \times 0.5 = 1.25$$ We want: $$P(X \leq 2) = \sum_{k=0}^2 \frac{1.25^k e^{-1.25}}{k!}$$ --- **Summary:** - $\lambda_{1\text{ min}} = 2.5$ - $P(X > 6) = 1 - \sum_{k=0}^6 \frac{2.5^k e^{-2.5}}{k!}$ - $\lambda_{8\text{ min}} = 20$ - $P(8 \leq X \leq 13) = \sum_{k=8}^{13} \frac{20^k e^{-20}}{k!}$ - $\lambda_{0.5\text{ min}} = 1.25$ - $P(X \leq 2) = \sum_{k=0}^2 \frac{1.25^k e^{-1.25}}{k!}$ These sums can be computed using a calculator or software for exact numerical values.