汽车随机到达一个码头，平均每 5 分钟到达一辆。渡轮将汽车从码头运送到一个岛上。渡轮一次最多可运送 3 辆车。当渡轮满载，或在到达后半小时，渡轮就会驶离，以时间较早者为准。

1. Identify the probability distribution and its mean for the 30-minute waiting period.

Let $X$ be the number of cars arriving at the jetty in 30 minutes.
The average rate is 1 car every 5 minutes, so the mean number of arrivals for 30 minutes is \lambda = rac{30}{5} = 6.
Thus, we model this with a Poisson distribution: $X \sim \text{Po}(6)$.

2. Calculate the probability that the ferry departs with exactly 2 cars.

The ferry departs with 2 cars only if exactly 2 cars arrive during the 30-minute interval (if 3 cars had arrived, it would have departed earlier as it would be full).
$P(X = 2) = \frac{e^{-6} 6^2}{2!} = 18e^{-6} \approx 0.044626$
Rounding to 3 significant figures, the probability is $0.0446$.

1. Identify the distribution for the number of cars arriving during the 30-minute maximum waiting time.

Let $X$ be the number of cars arriving at the jetty in 30 minutes. Since cars arrive randomly at an average rate of 1 every 5 minutes, the mean number of cars in 30 minutes is $\lambda = \frac{30}{5} = 6$.
Thus, $X \sim \text{Po}(6)$.

2. Calculate the probability that the ferry departs before it is full.

For the ferry to depart when it is still not full, it must wait the maximum of 30 minutes, which means fewer than 3 cars arrived in that time (i.e. $X < 3$).
$\begin{aligned} \text{P}(X \le 2) &= \text{P}(X = 0) + \text{P}(X = 1) + \text{P}(X = 2) \\ &= \text{e}^{-6} + 6\text{e}^{-6} + \frac{6^2}{2!}\text{e}^{-6} \\ &= 25\text{e}^{-6} \\ &\approx 0.0619688 \approx 0.0620 \end{aligned}$