一家制造公司生产医用口罩。这些医用口罩被随机装入每盒 50 个的包装盒中。平均而言，有 15% 的医用口罩是有缺陷的。一盒 50 个口罩中缺陷口罩的数量记为 $X$。 (i) 结合具体情境，陈述将一盒中缺陷医用口罩的数量用二项分布很好地进行建模所需的两个假设。 现在假设一盒中缺陷医用口罩的数量服从二项分布。 在随机抽取的一盒 50 个医用口罩中，求满足以下条件的概率：

1. State the assumption of independence in context

Whether any surgical mask in a box is defective is independent of whether any other surgical mask in the box is defective.

2. State the assumption of constant probability in context

The probability that any surgical mask in a box is defective is a constant $0.15$.

1. Identify the distribution and parameters of the random variable $X$.

Given $n = 50$ and $p = 0.15$:
$X \sim \text{B}(50, 0.15)$

2. Express 'more than 8' in terms of an inequality and use the complement rule.

$\text{P}(X > 8) = 1 - \text{P}(X \le 8)$

3. Evaluate the probability using a graphing calculator.

$\text{P}(X > 8) = 1 - 0.63466$
$= 0.36534$
$\approx 0.365 \text{ (to 3 s.f.)}$

1. Set up the conditional probability formula.

$\text{P}(X \leqslant 12 \mid X > 8) = \frac{\text{P}(X \leqslant 12 \cap X > 8)}{\text{P}(X > 8)}$
$= \frac{\text{P}(8 < X \leqslant 12)}{\text{P}(X > 8)}$

2. Express the numerator and denominator in terms of cumulative probabilities.

$= \frac{\text{P}(X \leqslant 12) - \text{P}(X \leqslant 8)}{1 - \text{P}(X \leqslant 8)}$

3. Evaluate the probabilities using a graphing calculator and find the final answer.

From $X \sim \text{B}(50, 0.15)$:
$\text{P}(X \leqslant 8) \approx 0.668065$
$\text{P}(X \leqslant 12) \approx 0.969941$

$\text{P}(X \leqslant 12 \mid X > 8) = \frac{0.969941 - 0.668065}{1 - 0.668065}$
$= \frac{0.301876}{0.331935} \approx 0.90944$
$\approx 0.909 \text{ (to 3 s.f.)}$

1. Identify the distribution of defective masks in a box and find the probability of at most 8 defective masks.

Let $X$ be the number of defective surgical masks in a box of $50$.
$X \sim \text{B}(50, 0.15)$

We require the probability that a randomly chosen box contains at most 8 defective masks:
$P(X \le 8) = 0.66651 \text{ (to 5 s.f.)}$

2. Define the distribution for the number of such boxes in a carton.

Let $Y$ be the number of boxes out of $24$ in a carton that contain at most $8$ defective masks.
$Y \sim \text{B}(24, 0.66651)$

3. Calculate the probability of finding at least 10 such boxes in a carton.

We need to find the probability that there are at least 10 such boxes:
$P(Y \ge 10) = 1 - P(Y \le 9)$
$P(Y \ge 10) = 1 - 0.003246 = 0.99675 \approx 0.997 \text{ (to 3 s.f.)}$

1. Set up and simplify the inequality comparing the probability of 1 defective mask to 0 defective masks.

Let $Y$ be the number of defective reusable masks in a packet of 10. Then $Y \sim \text{B}(10, p)$.
Since the unique modal number is 1, $P(Y=1) > P(Y=0)$.
$\binom{10}{1} p (1-p)^9 > \binom{10}{0} p^0 (1-p)^{10}$
$10p(1-p)^9 > (1-p)^{10}$

2. Solve the first inequality for $p$.

Since $p$ is a probability and $1$ is the mode, $0 < p < 1$, which means $(1-p)^9 > 0$. Dividing both sides by $(1-p)^9$:
$10p > 1 - p$
$11p > 1$
$p > \frac{1}{11}$

3. Set up and simplify the inequality comparing the probability of 1 defective mask to 2 defective masks.

For 1 to be the unique mode, we must also have $P(Y=1) > P(Y=2)$.
$\binom{10}{1} p (1-p)^9 > \binom{10}{2} p^2 (1-p)^8$
$10p(1-p)^9 > 45p^2(1-p)^8$

4. Solve the second inequality and combine the results to find the exact range of $p$.

Dividing both sides by $5p(1-p)^8$:
$2(1-p) > 9p$
$2 - 2p > 9p$
$11p < 2$
$p < \frac{2}{11}$
Combining both conditions, the exact range is:
$\frac{1}{11} < p < \frac{2}{11}$