In each batch of pralines that a chocolate factory produces, they are packed into boxes of 16. Due to the new health guidelines regarding sugar content, quality control tests are implemented and show that on average, the proportion of pralines that exceed the recommended sugar content is $p$. The number of pralines in a box that exceed the recommended sugar content is denoted by $X$.

1. State the assumption of independence in context

The event that a praline exceeds the recommended sugar content must be independent of whether any other praline exceeds the recommended sugar content.

2. State the assumption of constant probability in context

The probability that a praline exceeds the recommended sugar content must be constant for all pralines in the box.

1. Set up the conditions for $X=2$ to be the most probable value.

For $X=2$ to be the unique mode (the most probable number), its probability must be strictly greater than the probabilities of adjacent values, so we must have $\mathrm{P}(X=2) > \mathrm{P}(X=1)$ and $\mathrm{P}(X=2) > \mathrm{P}(X=3)$.

Using the binomial probability formula $\mathrm{P}(X=x) = \binom{n}{x}p^x(1-p)^{n-x}$ with $n=16$:
$\mathrm{P}(X=2) = \binom{16}{2}p^2(1-p)^{14} = 120p^2(1-p)^{14}$
$\mathrm{P}(X=1) = \binom{16}{1}p^1(1-p)^{15} = 16p(1-p)^{15}$
$\mathrm{P}(X=3) = \binom{16}{3}p^3(1-p)^{13} = 560p^3(1-p)^{13}$

2. Solve the inequalities to find the exact range of $p$.

From $\mathrm{P}(X=2) > \mathrm{P}(X=1)$:
$120p^2(1-p)^{14} > 16p(1-p)^{15}$
Since $p > 0$ and $1-p > 0$, we can divide both sides by $8p(1-p)^{14}$:
$15p > 2(1-p)$
$15p > 2 - 2p \implies 17p > 2 \implies p > \frac{2}{17}$

From $\mathrm{P}(X=2) > \mathrm{P}(X=3)$:
$120p^2(1-p)^{14} > 560p^3(1-p)^{13}$
Divide both sides by $40p^2(1-p)^{13}$:
$3(1-p) > 14p$
$3 - 3p > 14p \implies 17p < 3 \implies p < \frac{3}{17}$

Combining both conditions gives the exact range: $\frac{2}{17} < p < \frac{3}{17}$.
(Note: Strict inequalities are used since 2 is the uniquely "most probable" number.)

1. Calculate the probability of finding at most 1 praline exceeding the recommended sugar content in a box.

Let $X$ be the number of pralines in a box of 16 that exceed the recommended sugar content.
Given $X \sim \text{B}(16, 0.15)$,
$P(X \le 1) = P(X = 0) + P(X = 1)$
$= \binom{16}{0}(0.15)^0(0.85)^{16} + \binom{16}{1}(0.15)^1(0.85)^{15}$
$= 0.074253 + 0.209633 \approx 0.28389$

2. Calculate the probability of finding exactly 2 pralines exceeding the recommended sugar content in a box.

$P(X = 2) = \binom{16}{2}(0.15)^2(0.85)^{14}$
$= 120(0.0225)(0.102761) \approx 0.27745$

3. Calculate the total probability of the batch being accepted.

The batch is accepted if at most 1 praline exceeds in the first box ($X_1 \le 1$), OR if exactly 2 exceed in the first box ($X_1 = 2$) and at most 1 exceeds in the second box ($X_2 \le 1$).
$\text{P}(\text{accepted}) = P(X_1 \le 1) + P(X_1 = 2) \times P(X_2 \le 1)$
$= 0.28389 + (0.27745 \times 0.28389)$
$= 0.28389 + 0.078765$
$= 0.362655 \approx 0.363 \text{ (to 3 s.f.)}$

1. Determine the individual probability of exactly 2 pralines exceeding the limit in a single box.

Let $X$ be the number of pralines exceeding the recommended sugar content in a box. $X \sim \mathrm{B}(16, 0.15)$.
$\mathrm{P}(X = 2) = \binom{16}{2}(0.15)^2(0.85)^{14} = 0.27748$

2. Express the required conditional probability in terms of single-box probabilities.

Let $A$ be the event that the batch is accepted, and $B$ be the event that a second box is tested.
The batch is accepted ($A$) if the first box has $\le 1$ exceeding ($X_1 \le 1$), OR if the first has exactly 2 and the second has $\le 1$ ($X_1 = 2 \cap X_2 \le 1$).
A second box is tested and accepted ($B \cap A$) only in the latter case.
$\mathrm{P}(B \mid A) = \frac{\mathrm{P}(B \cap A)}{\mathrm{P}(A)} = \frac{\mathrm{P}(X_1 = 2)\mathrm{P}(X_2 \le 1)}{\mathrm{P}(X_1 \le 1) + \mathrm{P}(X_1 = 2)\mathrm{P}(X_2 \le 1)}$

3. Simplify the expression and evaluate the final probability.

Since the boxes are independent and identically distributed, $\mathrm{P}(X_1 \le 1) = \mathrm{P}(X_2 \le 1)$. We can factor out and cancel $\mathrm{P}(X \le 1)$ from the numerator and denominator:
$\mathrm{P}(B \mid A) = \frac{\mathrm{P}(X = 2)}{1 + \mathrm{P}(X = 2)}$
$= \frac{0.27748}{1 + 0.27748}$
$= 0.21721 \approx 0.217$ (to 3 s.f.)