At the drinks stall in a canteen, both hot and cold drinks are sold. The number of hot drinks sold per minute follows a Poisson distribution with mean 0.4. The number of cold drinks sold per minute follows an independent Poisson distribution with mean 0.5.

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.)