抛掷一枚特定硬币时获得正面的概率为 $p$。一个均匀的立方体骰子，其一个面写有数字“1”，两个面写有数字“2”，三个面写有数字“3”。 同时抛掷该硬币和骰子。随机变量 $X$ 定义如下： $\text{如果硬币显示正面，则 } X \text{ 为骰子点数的三倍。}$
$\text{如果硬币显示反面，则 } X \text{ 为骰子点数。}$

1. Determine the probabilities of obtaining each score on the biased die.

Let $D$ be the score on the die.
$\begin{aligned} \text{P}(D=1) &= \frac{1}{6} \\ \text{P}(D=2) &= \frac{2}{6} = \frac{1}{3} \\ \text{P}(D=3) &= \frac{3}{6} = \frac{1}{2} \end{aligned}$

2. Calculate the probability that $X = 3$ to show the given result.

Let $H$ and $T$ represent obtaining a head and tail respectively. Then $\text{P}(H) = p$ and $\text{P}(T) = 1-p$.
$X=3$ occurs either when a tail is thrown and the die score is 3, or when a head is thrown and the die score is 1.
$\begin{aligned} \text{P}(X=3) &= \text{P}(T)\text{P}(D=3) + \text{P}(H)\text{P}(D=1) \\ &= (1-p)\left(\frac{1}{2}\right) + p\left(\frac{1}{6}\right) \\ &= \frac{1}{2} - \frac{1}{2}p + \frac{1}{6}p \\ &= \frac{1}{2} - \frac{1}{3}p \quad \text{(Shown)} \end{aligned}$

3. Calculate the probabilities for all other possible values of $X$.

The other possible values of $X$ are $1$ and $2$ (from obtaining a tail), and $3\times 2=6$ and $3\times 3=9$ (from obtaining a head).
$\begin{aligned} \text{P}(X=1) &= \text{P}(T)\text{P}(D=1) = \frac{1}{6}(1-p) \\ \text{P}(X=2) &= \text{P}(T)\text{P}(D=2) = \frac{1}{3}(1-p) \\ \text{P}(X=6) &= \text{P}(H)\text{P}(D=2) = \frac{1}{3}p \\ \text{P}(X=9) &= \text{P}(H)\text{P}(D=3) = \frac{1}{2}p \end{aligned}$

4. Summarise the probability distribution in a table.

1. Use the probability distribution of $X$ from part (a) to set up the equation for $\text{E}(X)$.

Using the probabilities of $X$ derived in part (a):

• $\text{P}(X=1) = \frac{1-p}{6}$
• $\text{P}(X=2) = \frac{1-p}{3}$
• $\text{P}(X=3) = \frac{1}{2}-\frac{1}{3}p$
• $\text{P}(X=6) = \frac{p}{3}$
• $\text{P}(X=9) = \frac{p}{2}$

Given $\text{E}(X) = \sum x\text{P}(X=x) = 5$, we substitute the values:
$1\left(\frac{1-p}{6}\right) + 2\left(\frac{1-p}{3}\right) + 3\left(\frac{1}{2}-\frac{1}{3}p\right) + 6\left(\frac{p}{3}\right) + 9\left(\frac{p}{2}\right) = 5$

2. Expand and group the terms to simplify the equation.

$\frac{1-p}{6} + \frac{2-2p}{3} + \frac{3}{2} - p + 2p + \frac{9p}{2} = 5$

Grouping the constant terms and $p$ terms:
$\left( \frac{1}{6} + \frac{2}{3} + \frac{3}{2} \right) + p\left( -\frac{1}{6} - \frac{2}{3} - 1 + 2 + \frac{9}{2} \right) = 5$
$\left( \frac{1+4+9}{6} \right) + p\left( \frac{-1-4-6+12+27}{6} \right) = 5$
$\frac{14}{6} + \frac{28}{6}p = 5$
$\frac{7}{3} + \frac{14}{3}p = 5$

3. Solve the linear equation to find the exact value of $p$.

Multiply the entire equation by 3 to clear the denominators:
$7 + 14p = 15$
$14p = 8$
$p = \frac{8}{14} = \frac{4}{7}$

1. Find the value of $p$ using the expectation given in part (b).

From the probability distribution of $X$ derived in part (a):
$\text{E}(X) = 1\left(\frac{1-p}{6}\right) + 2\left(\frac{1-p}{3}\right) + 3\left(\frac{1}{2}-\frac{p}{3}\right) + 6\left(\frac{p}{3}\right) + 9\left(\frac{p}{2}\right) = 5$
$\frac{1-p}{6} + \frac{4-4p}{6} + \frac{9-6p}{6} + \frac{12p}{6} + \frac{27p}{6} = 5$
$\frac{14 + 28p}{6} = 5 \implies 14 + 28p = 30 \implies 28p = 16 \implies p = \frac{4}{7}$

2. Calculate $\text{E}(X^2)$ using the probability distribution and the found value of $p$.

$\text{E}(X^2) = 1^2\left(\frac{1-p}{6}\right) + 2^2\left(\frac{1-p}{3}\right) + 3^2\left(\frac{1}{2}-\frac{p}{3}\right) + 6^2\left(\frac{p}{3}\right) + 9^2\left(\frac{p}{2}\right)$
$\text{E}(X^2) = \frac{1-p}{6} + \frac{8-8p}{6} + \frac{27-18p}{6} + \frac{72p}{6} + \frac{243p}{6}$
$\text{E}(X^2) = \frac{36 + 288p}{6} = 6 + 48p$
Substitute $p = \frac{4}{7}$:
$\text{E}(X^2) = 6 + 48\left(\frac{4}{7}\right) = \frac{42}{7} + \frac{192}{7} = \frac{234}{7}$

3. Calculate the exact value of $\text{Var}(X)$ using the variance formula.

$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$
$\text{Var}(X) = \frac{234}{7} - 5^2$
$\text{Var}(X) = \frac{234}{7} - \frac{175}{7} = \frac{59}{7}$

1. Find the distribution of the difference in diameters and calculate the probability of a pair being acceptable.

Let $B$ and $H$ be the diameters of a randomly chosen bolt and hole respectively.
$B \sim \text{N}(2.65, 0.03^2)$
$H \sim \text{N}(2.72, 0.02^2)$

Let $Y = H - B$ be the difference in diameter.
$\text{E}(Y) = 2.72 - 2.65 = 0.07$
$\text{Var}(Y) = 0.02^2 + 0.03^2 = 0.0013$
$Y \sim \text{N}(0.07, 0.0013)$

A pair is acceptable if the hole is at least $0.02\text{ cm}$ larger but no more than $0.15\text{ cm}$ larger than the bolt:
$\text{P}(\text{acceptable}) = \text{P}(0.02 \le Y \le 0.15)$
$\text{P}(0.02 \le Y \le 0.15) \approx 0.903989$

2. Define the distribution for the number of acceptable pairs in a single box and find its mean and variance.

Let $X$ be the number of acceptable pairs in a box of 20.
$X \sim \text{B}(20, 0.903989)$

$\text{E}(X) = np = 20 \times 0.903989 = 18.07978$
$\text{Var}(X) = np(1-p) = 20 \times 0.903989 \times (1 - 0.903989) = 1.73586$

3. Apply the Central Limit Theorem to find the distribution of the average number of acceptable pairs in 50 boxes.

Let $\bar{X}$ be the average number of acceptable pairs in 50 boxes.
Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed.

$\bar{X} \sim \text{N}\left(\text{E}(X), \frac{\text{Var}(X)}{n}\right)$
$\bar{X} \sim \text{N}\left(18.07978, \frac{1.73586}{50}\right)$
$\bar{X} \sim \text{N}(18.07978, 0.034717)$

4. Calculate the probability that the average number is more than 18.

We need to find $\text{P}(\bar{X} > 18)$.

Using the normal distribution obtained:
$\text{P}(\bar{X} > 18) \approx 0.66576$
$\text{P}(\bar{X} > 18) = 0.666 \text{ (to 3 s.f.)}$