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 and its distribution.

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

2. Calculate the probability that at least 7 taxis arrive.

$\text{P}(T \ge 7) = 1 - \text{P}(T \le 6)$
$= 1 - 0.313374$
$= 0.686626$
$\approx 0.687 \text{ (to 3 s.f.)}$

1. Set up the conditional probability expression for the event.

Let $X$ and $Y$ be the number of Green Top and EZCab taxis arriving in a 10-minute period, respectively. Let $W = X + Y$ be the total number of taxis.
Given that $W \ge 7$, the event that 'all these taxis were EZCab taxis' means that $X = 0$ and $Y \ge 7$.
$P(X=0 \mid W \ge 7) = \frac{P(X=0 \text{ and } Y \ge 7)}{P(W \ge 7)}$

2. Calculate the numerator using the independence of $X$ and $Y$.

Since $X$ and $Y$ are independent:
$P(X=0 \text{ and } Y \ge 7) = P(X=0) \times P(Y \ge 7)$
With $X \sim \text{Po}(3)$ and $Y \sim \text{Po}(5)$:
$P(X=0) = e^{-3} \approx 0.049787$
$P(Y \ge 7) = 1 - P(Y \le 6) \approx 1 - 0.762183 = 0.237817$
$P(X=0 \text{ and } Y \ge 7) \approx 0.049787 \times 0.237817 \approx 0.011840$

3. Evaluate the final conditional probability using the result from part (i).

From part (i), $P(W \ge 7) \approx 0.686626$.
$\text{Probability} = \frac{0.011840}{0.686626} \approx 0.017244$
$\approx 0.0172 \text{ (to 3 s.f.)}$