某公司生产便携式扬声器。有些扬声器被发现是有缺陷的。每个纸箱装有 24 个便携式扬声器。

1. State the assumption regarding constant probability in context.

The probability of a portable speaker being faulty is constant.

2. State the assumption regarding independence in context.

Whether a portable speaker is faulty is independent of whether any other portable speaker is faulty.

1. Define the random variable and state its distribution.

Let $X$ be the number of faulty portable speakers in a carton.
Since each carton has 24 speakers and the probability of a faulty speaker is $2\%$, we have:
$X \sim \mathrm{B}(24, 0.02)$

2. Calculate the probability that there is at most one faulty speaker.

We need to find $\mathrm{P}(X \le 1)$.

1. Calculate the probability that a carton contains at most one faulty speaker (needed from part ii).

Let $X$ be the number of faulty speakers in a carton.
$X \sim \text{B}(24, 0.02)$
$\begin{aligned}\text{P}(X \le 1) &= \text{P}(X = 0) + \text{P}(X = 1) \\ &= \binom{24}{0}(0.02)^0(0.98)^{24} + \binom{24}{1}(0.02)^1(0.98)^{23} \\ &= 0.615847 + 0.301639 \\ &\approx 0.917486\end{aligned}$

2. Find the probability that there are no substandard cartons out of 50.

A carton is substandard if it has more than one faulty speaker ($X > 1$).
Thus, a carton is not substandard if $X \le 1$.
The probability of zero substandard cartons out of 50 is:
$\begin{aligned} \text{P}(\text{0 substandard}) &= [\text{P}(X \le 1)]^{50} \\ &= (0.917486)^{50} \\ &= 0.013496... \\ &\approx 0.0135 \text{ (to 3 s.f.)} \end{aligned}$

1. Define the random variable and its distribution for the number of faulty flash drives in a sample of 3.

Let $X$ be the number of faulty flash drives in a random sample of 3.
Since the status of each flash drive is independent, the number of faulty drives in any sample of 3 follows a binomial distribution:
$X \sim \mathrm{B}(3, p)$

2. Set up the probability equation for accepting a batch based on the quality control rules.

A batch is accepted in two mutually exclusive cases:

1. The first sample has 0 faulty drives ($X=0$).
2. The first sample has 1 faulty drive ($X=1$) and the second independent sample has 0 faulty drives ($X=0$).

$\mathrm{P}(\text{accept}) = \mathrm{P}(X=0) + \mathrm{P}(X=1) \times \mathrm{P}(X=0)$

3. Express the acceptance probability algebraically in terms of $p$.

Using the binomial probability formula:
$\mathrm{P}(X=0) = \binom{3}{0}p^0(1-p)^3 = (1-p)^3$
$\mathrm{P}(X=1) = \binom{3}{1}p^1(1-p)^2 = 3p(1-p)^2$

Substituting these into the equation:
$\mathrm{P}(\text{accept}) = (1-p)^3 + \left[3p(1-p)^2\right] \times (1-p)^3$
$\mathrm{P}(\text{accept}) = (1-p)^3 + 3p(1-p)^5$

4. Formulate the inequality and solve for the range of $p$ using a graphing calculator.

We require the probability of accepting a batch to be at least $95\%$ ($0.95$):
$(1-p)^3 + 3p(1-p)^5 \ge 0.95$
Using a graphing calculator to solve $(1-p)^3 + 3p(1-p)^5 = 0.95$, we find the critical value $p \approx 0.070348$.
Since $\mathrm{P}(\text{accept})$ decreases as the fault rate $p$ increases, the inequality holds for $p \le 0.070348$.
Given the problem's initial condition $0 < p < 1$, the required range is:
$0 < p \le 0.0703 \text{ (to 3 s.f.)}$