在食堂的饮料摊上，既销售热饮也销售冷饮。每分钟售出的热饮数量服从均值为 0.4 的泊松分布。每分钟售出的冷饮数量服从一个独立的、均值为 0.5 的泊松分布。

1. Determine the distribution of the total number of drinks sold in a 5-minute period.

Let $T$ be the total number of drinks sold in a 5-minute period.
The mean number of hot drinks sold in 5 minutes is $0.4 \times 5 = 2$.
The mean number of cold drinks sold in 5 minutes is $0.5 \times 5 = 2.5$.
Since the distributions are independent, the total number of drinks sold follows a Poisson distribution with mean $\lambda = 2 + 2.5 = 4.5$.
Thus, $T \sim \text{Po}(4.5)$.

2. Calculate the probability that more than 4 drinks are sold.

We need to find $\text{P}(T > 4)$.
Using the complement method:
$\text{P}(T > 4) = 1 - \text{P}(T \le 4)$
Evaluating this using a graphing calculator gives:
$\text{P}(T > 4) \approx 1 - 0.53210$
$= 0.46790$
$\approx 0.468 \text{ (to 3 s.f.)}$

1. Define the distributions for the number of hot, cold, and total drinks sold in a 5-minute period.

Let $X$ and $Y$ be the number of hot and cold drinks sold in a 5-minute period respectively.
$\lambda_X = 0.4 \times 5 = 2 \implies X \sim \text{Po}(2)$
$\lambda_Y = 0.5 \times 5 = 2.5 \implies Y \sim \text{Po}(2.5)$
Let $T$ be the total number of drinks sold, so $T = X + Y \sim \text{Po}(4.5)$.

2. Express the required probability as a conditional probability.

The condition "at most 4 drinks were sold" means $T \le 4$.
The event "they are all cold drinks" means $X = 0$ (no hot drinks were sold).
$\begin{aligned} \text{Required probability} &= P(X = 0 \mid T \le 4) \\ &= \frac{P(X = 0 \text{ and } T \le 4)}{P(T \le 4)} \\ &= \frac{P(X = 0 \text{ and } Y \le 4)}{P(T \le 4)} \end{aligned}$

3. Evaluate the probabilities using a graphing calculator to find the final answer.

Since $X$ and $Y$ are independent:
$P(X = 0 \text{ and } Y \le 4) = P(X = 0) P(Y \le 4)$
Using a graphing calculator:
$P(X = 0) \approx 0.135335$
$P(Y \le 4) \approx 0.891178$
$P(T \le 4) \approx 0.532104$
$\begin{aligned} \text{Required probability} &= \frac{0.135335 \times 0.891178}{0.532104} \\ &= 0.22666 \dots \\ &\approx 0.227 \text{ (to 3 s.f.)} \end{aligned}$

1. Define the binomial distribution for the number of periods where more than 4 drinks are sold.

Let $Y$ be the number of 5-minute periods where more than 4 drinks were sold.
Using the more precise probability from (i), $p \approx 0.46790$ (to 5 s.f.),
$Y \sim \text{B}(20, 0.46790)$

2. Formulate the probability inequality for the problem.

We are given that the probability of this occurring in at least $n$ periods is less than $0.4$.
$P(Y \ge n) < 0.4$
$1 - P(Y \le n-1) < 0.4$
$P(Y \le n-1) > 0.6$

3. Test values of $n$ using a graphing calculator to find the least integer.

Using a graphing calculator to generate values for $P(Y \ge n)$:
When $n=10$, $P(Y \ge 10) \approx 0.4126 \not< 0.4$
When $n=11$, $P(Y \ge 11) \approx 0.2242 < 0.4$

Thus, the least value of $n$ is $11$.

1. Find the mean number of hot and cold drinks sold in a 30-minute period.

Let $X$ and $Y$ be the number of hot and cold drinks sold in 30 minutes, respectively.
$X \sim \text{Po}(0.4 \times 30) \Rightarrow X \sim \text{Po}(12)$
$Y \sim \text{Po}(0.5 \times 30) \Rightarrow Y \sim \text{Po}(15)$

2. Determine the normal approximation for the difference in the number of drinks sold.

Since the means are both greater than 10, we can approximate their distributions using the Normal distribution.
$\text{E}(X - Y) = 12 - 15 = -3$
$\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) = 12 + 15 = 27$
Therefore, $X - Y \sim \text{N}(-3, 27)$ approximately.

3. Apply continuity correction for the probability calculation.

We need the probability that more hot drinks are sold than cold drinks, which is $P(X > Y)$ or $P(X - Y > 0)$.
Since $X$ and $Y$ represent discrete counts, $X - Y$ is an integer, so $X - Y \ge 1$.
Applying the continuity correction, we evaluate the continuous approximation at $> 0.5$, finding $P(X - Y > 0.5)$.

4. Calculate the required probability using the standard normal distribution.

$P(X - Y > 0.5) = P\left(Z > \frac{0.5 - (-3)}{\sqrt{27}}\right)$
$= P\left(Z > \frac{3.5}{\sqrt{27}}\right)$
$\approx P(Z > 0.67357)$
$\approx 0.250$ (to 3 s.f.)