Tetrahedral dice have four faces. Two fair tetrahedral dice, one red and one blue, have faces numbered 0, 1, 2, and 3 respectively. The dice are rolled, and the numbers faced down on the two dice are recorded. The random variable $T$ is defined as the score on the red die multiplied by the score on the blue die.

1. Construct a table of outcomes to find all possible values of $T$.

The possible scores on each die are $0, 1, 2, 3$. There are $4 \times 4 = 16$ equally likely outcomes for the pair of dice.
The product of the scores, $T$, can be represented in a table of outcomes:

2. Calculate the probability for each value of $T$.

By counting the frequencies of each possible value of $T$ out of the $16$ total outcomes, we have:
$P(T=0) = \frac{7}{16}$
$P(T=1) = \frac{1}{16}$
$P(T=2) = \frac{2}{16} = \frac{1}{8}$
$P(T=3) = \frac{2}{16} = \frac{1}{8}$
$P(T=4) = \frac{1}{16}$
$P(T=6) = \frac{2}{16} = \frac{1}{8}$
$P(T=9) = \frac{1}{16}$

3. Tabulate the probability distribution of $T$.

Summarising the probabilities into a formal distribution table:

1. Determine the probability distribution of $T$ (from part (i)) and calculate the expected value $E(T)$.

The possible outcomes for $T = X \times Y$ where $X, Y \in \{0, 1, 2, 3\}$ give the following probability distribution:
$P(T=0) = \frac{7}{16}$, $P(T=1) = \frac{1}{16}$, $P(T=2) = \frac{2}{16}$, $P(T=3) = \frac{2}{16}$, $P(T=4) = \frac{1}{16}$, $P(T=6) = \frac{2}{16}$, $P(T=9) = \frac{1}{16}$.

$\begin{aligned} E(T) &= \sum t P(T=t) \\ &= 0\left(\frac{7}{16}\right) + 1\left(\frac{1}{16}\right) + 2\left(\frac{2}{16}\right) + 3\left(\frac{2}{16}\right) + 4\left(\frac{1}{16}\right) + 6\left(\frac{2}{16}\right) + 9\left(\frac{1}{16}\right) \\ &= \frac{0 + 1 + 4 + 6 + 4 + 12 + 9}{16} \\ &= \frac{36}{16} \\ &= \frac{9}{4} \end{aligned}$

2. Calculate $E(T^2)$ and apply the variance formula $\mathrm{Var}(T) = E(T^2) - [E(T)]^2$ to show the required result.

$\begin{aligned} E(T^2) &= \sum t^2 P(T=t) \\ &= 0^2\left(\frac{7}{16}\right) + 1^2\left(\frac{1}{16}\right) + 2^2\left(\frac{2}{16}\right) + 3^2\left(\frac{2}{16}\right) + 4^2\left(\frac{1}{16}\right) + 6^2\left(\frac{2}{16}\right) + 9^2\left(\frac{1}{16}\right) \\ &= \frac{0 + 1 + 8 + 18 + 16 + 72 + 81}{16} \\ &= \frac{196}{16} \\ &= \frac{49}{4} \end{aligned}$

$\begin{aligned} \mathrm{Var}(T) &= E(T^2) - [E(T)]^2 \\ &= \frac{49}{4} - \left(\frac{9}{4}\right)^2 \\ &= \frac{196}{16} - \frac{81}{16} \\ &= \frac{115}{16} \end{aligned}$

1. Substitute the values of $\mu$ and $\sigma$ to form an inequality for $T$.

From part (ii), we have $E(T) = \mu = \frac{9}{4} = 2.25$ and $\mathrm{Var}(T) = \frac{115}{16}$.
Thus, $\sigma = \sqrt{\frac{115}{16}} = \frac{\sqrt{115}}{4} \approx 2.681$.
The expression $|T - 2\mu| > \sigma$ becomes:
$|T - 2(2.25)| > 2.681$
$|T - 4.5| > 2.681$

2. Solve the modulus inequality to find the range of valid values for $T$.

$T - 4.5 < -2.681 \quad \text{or} \quad T - 4.5 > 2.681$
$T < 1.819 \quad \text{or} \quad T > 7.181$

3. Identify the values of $T$ from its probability distribution that satisfy the inequality.

The possible values of $T$ (the product of the scores on the two dice) are $0, 1, 2, 3, 4, 6,$ and $9$.
The values of $T$ that satisfy $T < 1.819$ or $T > 7.181$ are $0, 1,$ and $9$.

4. Calculate the total probability for the identified values of $T$.

$\mathrm{P}(|T - 2\mu| > \sigma) = \mathrm{P}(T = 0) + \mathrm{P}(T = 1) + \mathrm{P}(T = 9)$
Using the probabilities of outcomes:
$= \frac{7}{16} + \frac{1}{16} + \frac{1}{16}$
$= \frac{9}{16}$