Cars arrive at a jetty randomly at an average rate of one every 5 minutes. From the jetty, a ferry transports the cars to an island. The ferry can transport at most 3 cars at a time. The ferry departs the jetty when it is full or half an hour after its arrival whichever is sooner.

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. $X < 3$).
$\begin{aligned} \text{P}(X \le 2) &= \text{P}(X = 0) + \text{P}(X = 1) + \text{P}(X = 2) \\ &= \text{e}^{-6} + 6\text{e}^{-6} + \frac{6^2}{2!}\text{e}^{-6} \\ &= 25\text{e}^{-6} \\ &\approx 0.0619688 \approx 0.0620 \end{aligned}$