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. For each batch of pralines produced, the chocolate factory implements the following quality control system by testing the first random sample of 16 pralines in a box:

• If there are at most 1 praline that exceed the recommended sugar content in a box, the batch is accepted.
• If there are 3 or more pralines that exceed the recommended sugar content in a box, the batch is rejected.
• If there are 2 pralines that exceed the recommended sugar content in a box, a second box of 16 pralines is tested. The batch will be accepted if there are at most 1 praline that exceed the recommended sugar content in the second box, otherwise the batch will be rejected.

It is given that $p = 0.15.

1. State the independence assumption in context

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

2. State the constant probability assumption in context

The probability of a praline exceeding the recommended sugar content is constant for all pralines in the box.

1. Formulate the conditions for $X=2$ to be the unique most probable outcome.

For $X=2$ to be the most probable number of successes, its probability must be strictly greater than its adjacent probabilities:
$P(X=2) > P(X=1)$
$P(X=2) > P(X=3)$

2. Set up and solve the inequality $P(X=2) > P(X=1)$ for $p$.

$\binom{16}{2}p^2(1-p)^{14} > \binom{16}{1}p(1-p)^{15}$
$120p^2(1-p)^{14} > 16p(1-p)^{15}$
Since $0 < p < 1$, we can divide both sides by $16p(1-p)^{14}$:
$\frac{15}{2}p > 1-p$
$15p > 2 - 2p$
$17p > 2 \implies p > \frac{2}{17}$

3. Set up and solve the inequality $P(X=2) > P(X=3)$ and combine the results.

$\binom{16}{2}p^2(1-p)^{14} > \binom{16}{3}p^3(1-p)^{13}$
$120p^2(1-p)^{14} > 560p^3(1-p)^{13}$
Divide both sides by $40p^2(1-p)^{13}$:
$3(1-p) > 14p$
$3 - 3p > 14p$
$17p < 3 \implies p < \frac{3}{17}$
Combining the two inequalities gives the exact range for $p$:
$\frac{2}{17} < p < \frac{3}{17}$

1. Formulate the total probability of accepting the batch

Let $X \sim \mathrm{B}(16, 0.15)$.

The batch is accepted if $X \le 1$ in the first box, or if $X = 2$ in the first box and $X \le 1$ in the second box.
$\mathrm{P}(\text{accepted}) = \mathrm{P}(X \le 1) + \mathrm{P}(X = 2) \times \mathrm{P}(X \le 1)$

2. Evaluate the probability using the binomial distribution

Using a graphing calculator:
$\mathrm{P}(X \le 1) \approx 0.28391$
$\mathrm{P}(X = 2) \approx 0.27748$

$\mathrm{P}(\text{accepted}) = 0.283907 + 0.277484 \times 0.283907$
$= 0.362687$
$\approx 0.363 \text{ (to 3 s.f.)}$

1. Set up the conditional probability using events from the problem

Let $A$ be the event that the batch is accepted, and $B$ be the event that a second box is tested.
$P(B \mid A) = \frac{P(B \cap A)}{P(A)}$
From the context, the batch is accepted after a second box is tested if the first box has exactly 2 defectives and the second box has $\le 1$ defective.
$P(B \cap A) = P(X_1 = 2) \times P(X_2 \le 1)$
The total probability of acceptance is:
$P(A) = P(X_1 \le 1) + P(X_1 = 2) \times P(X_2 \le 1)$

2. Substitute the binomial probabilities and calculate the final probability

Using $X \sim \text{B}(16, 0.15)$:
$P(X \le 1) \approx 0.28391$
$P(X = 2) \approx 0.27748$

Substituting these values:
$P(B \mid A) = \frac{0.27748 \times 0.28391}{0.28391 + (0.27748 \times 0.28391)}$
$= \frac{0.078780}{0.36269}$
$= 0.21721 \dots$
$\approx 0.217 \text{ (3 s.f.)}$