A game is played in which a tetrahedral die is thrown. The sides of the die are numbered 1, 2, 3, 4. Whatever number on the face that the die land on, that many number of fair coins are tossed. A player's final score $X$, is the number of coins showing heads.

1. Establish the probability model for the number of heads based on the die outcome.

Let $D$ be the outcome on the tetrahedral die, so $\mathrm{P}(D=d) = \frac{1}{4}$ for $d = 1, 2, 3, 4$.
Given $D=d$, $d$ fair coins are tossed. The number of heads $X$ follows a Binomial distribution:
$X|D=d \sim \mathrm{B}(d, 0.5)$

2. Calculate $\mathrm{P}(X=0)$ and $\mathrm{P}(X=1)$ using total probability.

Using $\mathrm{P}(X=x) = \sum_{d=1}^4 \mathrm{P}(D=d)\mathrm{P}(X=x|D=d)$:

3. Calculate $\mathrm{P}(X=2)$, $\mathrm{P}(X=3)$ and $\mathrm{P}(X=4)$.

For $X=2$, note that $D$ must be at least $2$:

For $X=3$, $D$ must be at least $3$:

For $X=4$, $D$ must be $4$:
$\mathrm{P}(X=4) = \frac{1}{4}\left[ 0 + 0 + 0 + \binom{4}{4}\left(\frac{1}{2}\right)^4 \right] = \frac{1}{4}\left(\frac{1}{16}\right) = \frac{1}{64}$

4. Summarize the probabilities into a distribution table.

The probability distribution table for $X$ is:

1. Establish the probability distribution of $X$ as required from part (i).

Given that the die lands on $n \in \{1, 2, 3, 4\}$ with $P(N=n) = \frac{1}{4}$, the number of heads $X$ follows a binomial distribution $B(n, 0.5)$ for a given $n$.
By the law of total probability, $P(X=x) = \sum_{n=1}^{4} P(N=n) P(X=x \mid N=n)$:
$P(X=0) = \frac{1}{4} \left[ \binom{1}{0}0.5^1 + \binom{2}{0}0.5^2 + \binom{3}{0}0.5^3 + \binom{4}{0}0.5^4 \right] = \frac{15}{64}$
$P(X=1) = \frac{1}{4} \left[ \binom{1}{1}0.5^1 + \binom{2}{1}0.5^2 + \binom{3}{1}0.5^3 + \binom{4}{1}0.5^4 \right] = \frac{26}{64}$
$P(X=2) = \frac{1}{4} \left[ 0 + \binom{2}{2}0.5^2 + \binom{3}{2}0.5^3 + \binom{4}{2}0.5^4 \right] = \frac{16}{64}$
$P(X=3) = \frac{1}{4} \left[ 0 + 0 + \binom{3}{3}0.5^3 + \binom{4}{3}0.5^4 \right] = \frac{6}{64}$
$P(X=4) = \frac{1}{4} \left[ 0 + 0 + 0 + \binom{4}{4}0.5^4 \right] = \frac{1}{64}$

2. Calculate the expected value $\text{E}(X)$.

Using the expected value formula $\text{E}(X) = \sum x P(X=x)$:
$\text{E}(X) = 0\left(\frac{15}{64}\right) + 1\left(\frac{26}{64}\right) + 2\left(\frac{16}{64}\right) + 3\left(\frac{6}{64}\right) + 4\left(\frac{1}{64}\right)$
$\text{E}(X) = \frac{0 + 26 + 32 + 18 + 4}{64} = \frac{80}{64} = \frac{5}{4}$

3. Calculate the variance $\text{Var}(X)$.

First, calculate $\text{E}(X^2) = \sum x^2 P(X=x)$:
$\text{E}(X^2) = 0^2\left(\frac{15}{64}\right) + 1^2\left(\frac{26}{64}\right) + 2^2\left(\frac{16}{64}\right) + 3^2\left(\frac{6}{64}\right) + 4^2\left(\frac{1}{64}\right)$
$\text{E}(X^2) = \frac{0 + 26 + 64 + 54 + 16}{64} = \frac{160}{64} = \frac{5}{2}$
Then, use the variance formula $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$:
$\text{Var}(X) = \frac{5}{2} - \left(\frac{5}{4}\right)^2 = \frac{40}{16} - \frac{25}{16} = \frac{15}{16}$