一个店主收到了一批文具，包括笔记本、钢笔和修正带。历史数据表明，1%的笔记本、2%的钢笔和4%的修正带是有缺陷的。笔记本、钢笔和修正带的质量是相互独立的。店主决定将这些文具打包成180个包装出售，每个包装包含一本笔记本、两支钢笔和一卷修正带。如果这四个文具中的任何一个是缺陷品，则该包装被视为不合格。

1. Calculate the probability that all items in a randomly selected packet are satisfactory.

The probability that each item is satisfactory is the complement of it being defective:
$P(\text{satisfactory notebook}) = 1 - 0.01 = 0.99$
$P(\text{satisfactory pen}) = 1 - 0.02 = 0.98$
$P(\text{satisfactory tape}) = 1 - 0.04 = 0.96$

Since the items are independent and a packet contains 1 notebook, 2 pens, and 1 correction tape:
$P(\text{all items satisfactory}) = 0.99 \times (0.98)^2 \times 0.96$
$= 0.91276416$

2. Use the complement rule to find the probability of an unsatisfactory packet and round to 3 significant figures.

A packet is unsatisfactory if at least one item is defective, which is the complement of all items being satisfactory.
$P(\text{unsatisfactory packet}) = 1 - P(\text{all items satisfactory})$
$= 1 - 0.91276416$
$= 0.08723584$
$\approx 0.0872 \text{ (correct to 3 s.f.)}$

1. Define the distribution of $X$ and express the required probability mathematically.

Using the exact unrounded probability from (i), let $p = 1 - (0.99)(0.98)^2(0.96) = 0.08723584$.
Given that $X \sim \text{B}(180, p)$.
We need to find the probability of at least 5 but less than 10 packets, which is $P(5 \le X < 10)$.
Since $X$ is a discrete random variable,
$P(5 \le X < 10) = P(5 \le X \le 9)$

2. Evaluate the probability using a graphing calculator.

$P(5 \le X \le 9) = P(X \le 9) - P(X \le 4)$
Using a graphing calculator:
$P(X \le 9) \approx 0.043315$
$P(X \le 4) \approx 0.000334$
$P(5 \le X \le 9) = 0.043315 - 0.000334$
$= 0.042981$
$\approx 0.0430 \text{ (to 3 s.f.)}$

1. Set up the probability inequality based on the given condition.

We are given that $\text{P}(X > r) \le 0.12$.
Using the complement, this can be rewritten as:
$1 - \text{P}(X \le r) \le 0.12$
$\text{P}(X \le r) \ge 0.88$

2. Use a graphing calculator to test consecutive values of $r$.

Using the exact unrounded probability of an unsatisfactory packet from (i), $p = 0.087236$, we have $X \sim \text{B}(180, 0.087236)$.
Using the table function on a graphing calculator:
When $r = 19$, $\text{P}(X \le 19) \approx 0.8437$, so $\text{P}(X > 19) \approx 0.156 > 0.12$
When $r = 20$, $\text{P}(X \le 20) \approx 0.8919$, so $\text{P}(X > 20) \approx 0.108 \le 0.12$

3. State the least value of $r$.

From the calculated probabilities, the condition $\text{P}(X > r) \le 0.12$ is first satisfied when $r = 20$.

1. State the method of selection to obtain a simple random sample.

Assign a unique number from $1$ to $180$ to each of the $180$ packets. Use a random number generator or calculator to generate $9$ unique random integers between $1$ and $180$. Select the $9$ packets corresponding to these numbers.

2. Provide the reason for choosing this selection method.

This method ensures that every packet (and every possible sample of size 9) has an equal chance of being selected, thus preventing bias.

1. State the probability of a packet being unsatisfactory and formulate the probability for the specific sequence of selections.

Let $p$ be the exact probability of a packet being unsatisfactory. As shown from part (i), $p = 1 - (0.99)(0.98)^2(0.96) = 0.08723584$.

For the 9th packet selected to be exactly the 3rd unsatisfactory packet, there must be exactly 2 unsatisfactory packets among the first 8 packets selected, and the 9th packet must be unsatisfactory.
$\text{Probability} = \left[ \binom{8}{2} p^2 (1-p)^6 \right] \times p$

2. Evaluate the exact expression and round to 3 significant figures.

$\text{Probability} = 28 \times (0.08723584)^3 \times (1 - 0.08723584)^6$
$= 28 \times 0.000663853 \dots \times 0.577937 \dots$
$= 0.0107426 \dots$
$\approx 0.0107 \text{ (to 3 s.f.)}$