The probability of obtaining a head when a particular coin is tossed is $p$. A fair cubical die has the number '1' on one face, number '2' on two faces and number '3' on three faces. The coin and die are thrown simultaneously. The random variable $X$ is defined as follows. $\text{If the coin shows a head, then } X \text{ is thrice the score on the die.}$
$\text{If the coin shows a tail, then } X \text{ is the score on the die.}$

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