安东尼用一个均匀的四面骰子玩游戏，骰子的四个面上分别印有数字 1、2、3 和 4。安东尼掷骰子，如果结果是质数，他将赢得与该数字等额的美元。相反，如果结果是非质数，安东尼将失去与该数字等额的美元。随机变量 $X$ 表示安东尼在一次掷骰子中赢取的金额。

1. Determine the possible values of $X$ and their probabilities, and construct the probability distribution table.

The possible outcomes of the die are 1, 2, 3, and 4.

• The prime numbers are 2 and 3, corresponding to wins ($X = 2$ and $X = 3$).
• The non-prime numbers are 1 and 4, corresponding to losses ($X = -1$ and $X = -4$).

Since the four-sided die is fair, each outcome has a probability of $\frac{1}{4}$.

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

$\text{E}(X) = \sum x \, P(X=x)$
$\text{E}(X) = (-4)\left(\frac{1}{4}\right) + (-1)\left(\frac{1}{4}\right) + (2)\left(\frac{1}{4}\right) + (3)\left(\frac{1}{4}\right)$
$\text{E}(X) = -1 - 0.25 + 0.5 + 0.75$
$\text{E}(X) = 0$

1. Calculate $\text{E}(X^2)$ based on the probability distribution from part (a)

From part (a), the values of $X$ are $-4$, $-1$, $2$, and $3$, each with probability $\frac{1}{4}$.

$\text{E}(X^2) = \sum x^2 \text{P}(X=x)$
$\text{E}(X^2) = \frac{1}{4}((-4)^2 + (-1)^2 + 2^2 + 3^2)$
$\text{E}(X^2) = \frac{1}{4}(16 + 1 + 4 + 9)$
$\text{E}(X^2) = \frac{1}{4}(30) = 7.5$

2. Calculate the variance using $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$

Using $\text{E}(X) = 0$ found in part (a):
$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$
$\text{Var}(X) = 7.5 - 0^2$
$\text{Var}(X) = 7.5$

1. Identify the values of $X$ and construct the table of outcomes for $|X_1 + X_2|$.

From part (a), the possible amounts won are $2$ and $3$ (prime numbers), and the possible amounts lost are $-1$ and $-4$ (non-prime numbers). Thus, the possible values of $X$ are $-4, -1, 2, \text{ and } 3$.

Using these values, we draw a table of outcomes for $|X_1 + X_2|$:

2. Calculate the expected value $\text{E}(|X_1 + X_2|)$.

Since the 4-sided die is fair, each of the $16$ outcomes in the table is equally likely and occurs with a probability of $\frac{1}{16}$.

To find the expected value, we sum all the values in the table and multiply by $\frac{1}{16}$:

1. Express the probability of the other equally likely outcomes in terms of $p$.

Let the probability of obtaining 2, 4, 6 or 7 be $q$.
Since the sum of all probabilities must be 1:
$p + 2p + 4q = 1$
$4q = 1 - 3p \implies q = \frac{1 - 3p}{4}$

2. Formulate the expected winnings in terms of $p$ and $q$.

Let $Y$ be the random variable for the amount of money won.
Identify the prime outcomes (winnings) as 2, 3, 5, 7 and non-prime outcomes (losses) as 4, 6.
$\text{E}(Y) = 2(q) + 3(p) - 4(q) + 5(2p) - 6(q) + 7(q)$
$\text{E}(Y) = 13p + q(2 - 4 - 6 + 7)$
$\text{E}(Y) = 13p - q$

3. Equate the expected winnings to zero for a fair game and solve for $p$.

For a game to be fair, the expected winnings must be zero, so $\text{E}(Y) = 0$.
$13p - q = 0 \implies q = 13p$
Substitute $q = \frac{1 - 3p}{4}$:
$13p = \frac{1 - 3p}{4}$
$52p = 1 - 3p$
$55p = 1 \implies p = \frac{1}{55}$