A shop sells apples in bags of 12. The average number of unripe apples in a bag is 2.16.

1. State the assumption regarding constant probability.

The probability that any individual apple is unripe is constant.

2. State the assumption regarding independence.

The event that an apple is unripe is independent of the ripeness of any other apple in the bag.

1. Find the probability $p$ of a randomly selected apple being unripe using the given mean.

$\text{E}(X) = np$
$12p = 2.16$
$p = 0.18$

2. Define the binomial random variable and formulate the required probability.

Let $X$ be the number of unripe apples in a bag of 12.
$X \sim \text{B}(12, 0.18)$
We require the probability that more than 3 apples are unripe ($X > 3$):
$P(X > 3) = 1 - P(X \le 3)$

3. Evaluate the probability using a graphing calculator.

$P(X > 3) = 1 - 0.844813$
$= 0.155187$
$\approx 0.155 \text{ (to 3 s.f.)}$

1. Find the probability of a bag being sold at a reduced price (result needed from part (b)).

Let $X$ be the number of unripe apples in a bag of $12$.
$E(X) = 12p = 2.16 \implies p = 0.18$
$\therefore X \sim \text{B}(12, 0.18)$
$P(\text{reduced price}) = P(X > 3) = 1 - P(X \le 3) \approx 0.15512$

2. Define the binomial distribution for the number of reduced price bags and find the required probability.

Let $Y$ be the number of bags sold at a reduced price out of $20$.
$Y \sim \text{B}(20, 0.15512)$
We want $P(Y < 4)$:
$P(Y < 4) = P(Y \le 3) \approx 0.62179 \approx 0.622 \text{ (to 3 s.f.)}$

1. Set up the conditions for the mode of a binomial distribution

Let $Y$ be the number of unripe oranges in a bag of $n$ oranges.
Assume $Y \sim \mathrm{B}(n, 0.16)$.
Since the modal number of unripe oranges is 2, the probability of getting exactly 2 unripe oranges must be greater than or equal to the probability of getting 1 or 3 unripe oranges.
Thus, we must have $\mathrm{P}(Y=2) \ge \mathrm{P}(Y=1)$ and $\mathrm{P}(Y=2) \ge \mathrm{P}(Y=3)$.

2. Solve the inequality $\mathrm{P}(Y=2) \ge \mathrm{P}(Y=1)$ for $n$

$\binom{n}{2} (0.16)^2 (0.84)^{n-2} \ge \binom{n}{1} (0.16)^1 (0.84)^{n-1}$
$\frac{\frac{n(n-1)}{2}}{n} \left( \frac{0.16}{0.84} \right) \ge 1$
$\frac{n-1}{2} \left( \frac{4}{21} \right) \ge 1$
$n-1 \ge \frac{21}{2} = 10.5$
$n \ge 11.5$

3. Solve the inequality $\mathrm{P}(Y=2) \ge \mathrm{P}(Y=3)$ for $n$

$\binom{n}{2} (0.16)^2 (0.84)^{n-2} \ge \binom{n}{3} (0.16)^3 (0.84)^{n-3}$
$\frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)}{6}} \left( \frac{0.84}{0.16} \right) \ge 1$
$\frac{3}{n-2} \left( \frac{21}{4} \right) \ge 1$
$\frac{63}{4} \ge n-2$
$15.75 \ge n-2 \implies n \le 17.75$

4. Determine the set of possible integer values for $n$

Combining the results from both inequalities, we have:
$11.5 \le n \le 17.75$
Since $n$ represents the number of oranges in a bag, it must be an integer.
The possible integer values are $12, 13, 14, 15, 16, 17$.
The set of possible values is $\{12, 13, 14, 15, 16, 17\}$.