The number of Green Top taxis and EZCab taxis arriving in a randomly chosen 10-minute period at the airport taxi bay may be assumed to be independent Poisson variables with mean 3 and 5 respectively.

1. Define the combined random variable for the total number of taxis.

Let $X$ be the number of Green Top taxis and $Y$ be the number of EZCab taxis arriving in a 10-minute period.
$X \sim \text{Po}(3)$ and $Y \sim \text{Po}(5)$.
Let $T = X + Y$ be the total number of taxis.
Since $X$ and $Y$ are independent, $T \sim \text{Po}(3 + 5)$, so $T \sim \text{Po}(8)$.

2. Set up the probability expression using the complement rule.

We want to find the probability of at least 7 taxis arriving, which is $\text{P}(T \ge 7)$.
$\text{P}(T \ge 7) = 1 - \text{P}(T \le 6)$

3. Evaluate the probability to show the required result.

Using a graphing calculator to evaluate the Poisson cumulative distribution function:
$\text{P}(T \le 6) \approx 0.31337$
$\text{P}(T \ge 7) = 1 - 0.31337 = 0.68663$
$\text{P}(T \ge 7) \approx 0.687 \text{ (to 3 s.f.)}$

1. Define variables and set up the conditional probability statement.

Let $X \sim \text{Po}(3)$ be the number of Green Top taxis, $Y \sim \text{Po}(5)$ be the number of EZCab taxis, and $T = X + Y \sim \text{Po}(8)$ be the total number of taxis in a 10-minute period.

We want the probability that all taxis are EZCab given that at least 7 taxis arrived. For all taxis to be EZCab, the number of Green Top taxis must be zero ($X = 0$). If $X = 0$ and $T \ge 7$, then $Y$ must be $\ge 7$.

$\text{Required probability} = \text{P}(X = 0 \text{ and } Y \ge 7 \mid T \ge 7)$
$= \frac{\text{P}(X = 0 \text{ and } Y \ge 7)}{\text{P}(T \ge 7)}$

2. Calculate the probability of the intersection using independence.

Since the arrivals of Green Top and EZCab taxis are independent:
$\text{P}(X = 0 \text{ and } Y \ge 7) = \text{P}(X = 0) \times \text{P}(Y \ge 7)$

Using the Poisson formulas or a graphing calculator:
$\text{P}(X = 0) = e^{-3} \approx 0.049787$
$\text{P}(Y \ge 7) = 1 - \text{P}(Y \le 6) \approx 1 - 0.762183 = 0.237817$

$\text{P}(X = 0 \text{ and } Y \ge 7) \approx 0.049787 \times 0.237817 \approx 0.011840$

3. Evaluate the final conditional probability.

Using the unrounded probability from part (i), $\text{P}(T \ge 7) = 1 - \text{P}(T \le 6) \approx 0.686626$.

$\text{P}(\text{all EZCab} \mid T \ge 7) = \frac{0.011840}{0.686626} \approx 0.017244$

Rounding to 3 significant figures, we get $0.0172$.