A manufacturing company produces surgical masks. The surgical masks are randomly packed into boxes of 50. On average, 15% of the surgical masks are defective. The number of surgical masks that are defective in a box of 50 pieces is denoted by $X$. (i) State, in context, two assumptions needed for the number of defective surgical masks in a box to be well modelled by a binomial distribution. Assume now that the number of defective surgical masks in a box has a binomial distribution. In a randomly chosen box of 50 surgical masks, find the probability that there are

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:
$\frac{1}{11} < p < \frac{2}{11}$