巧克力工厂生产的每批夹心巧克力都以每盒16颗进行包装。根据关于含糖量的新健康指南，实施了质量控制检测，结果表明平均而言，含糖量超过推荐标准的夹心巧克力比例为 $p$。一盒中含糖量超过推荐标准的夹心巧克力数量记为 $X. 对于生产的每一批夹心巧克力，巧克力工厂通过测试一盒中由16颗巧克力组成的第一个随机样本，实施以下质量控制系统：

• 如果一盒中最多有 1 颗夹心巧克力的含糖量超过推荐标准，则该批次被接收。
• 如果一盒中有 3 颗或更多夹心巧克力的含糖量超过推荐标准，则该批次被拒收。
• 如果一盒中有 2 颗夹心巧克力的含糖量超过推荐标准，则测试第二盒包含 16 颗夹心巧克力的样本。如果第二盒中最多有 1 颗夹心巧克力的含糖量超过推荐标准，则该批次被接收；否则，该批次被拒收。

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