Here is a clear, educational, and well-structured solution to the problem.

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1. Key Concepts and Formulas

To solve this problem, we will utilize fundamental concepts from probability and statistics:

• Expected Value ($E[X]$ or Mean): For a discrete random variable $X$ with possible values $x_i$ and probabilities $P(X=x_i)$, its expected value is given by: $E[X] = \sum_{i} x_i P(X=x_i)$
• Variance ($Var[X]$): This measures the spread of the values of a random variable around its expected value. It is defined by the formula: $Var[X] = E[X^2] - (E[X])^2$ where $E[X^2] = \sum_{i} x_i^2 P(X=x_i)$.
• Sum of Positive Divisors ($\sigma(n)$): For a positive integer $n$ with prime factorization $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, the sum of its positive divisors is: $\sigma(n) = (1+p_1+\cdots+p_1^{e_1})(1+p_2+\cdots+p_2^{e_2})\cdots(1+p_k+\cdots+p_k^{e_k})$ This can also be expressed as $\sigma(n) = \left(\frac{p_1^{e_1+1}-1}{p_1-1}\right) \left(\frac{p_2^{e_2+1}-1}{p_2-1}\right) \cdots \left(\frac{p_k^{e_k+1}-1}{p_k-1}\right)$.

2. Step-by-Step Solution

Step 1: Define the Random Variable and its Sample Space.

• What we are doing: We identify the random variables involved and determine the range of values for our target random variable, $X = \alpha - \beta$.
• Why we are doing this: Clearly defining the random variable and its possible outcomes is the first crucial step to constructing its probability distribution.
• Calculations: Let $\alpha$ be the outcome on die A and $\beta$ be the outcome on die B. The possible values for $\alpha$ and $\beta$ are $\{1, 2, 3, 4, 5, 6\}$. Since two dice are rolled, there are $6 \times 6 = 36$ equally likely outcomes $(\alpha, \beta)$. Our random variable is $X = \alpha - \beta$. The minimum value of $X$ occurs when $\alpha=1$ and $\beta=6$, so $X_{min} = 1 - 6 = -5$. The maximum value of $X$ occurs when $\alpha=6$ and $\beta=1$, so $X_{max} = 6 - 1 = 5$. Thus, the possible values for $X$ are $\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$.

Step 2: Construct the Probability Distribution of $X = \alpha - \beta$.

• What we are doing: We systematically determine the number of ways each value of $X$ can occur and then calculate its probability.
• Why we are doing this: The expected value and variance calculations depend directly on this probability distribution.
• Calculations: We count the number of pairs $(\alpha, \beta)$ such that $\alpha - \beta = x_i$ for each $x_i \in \{-5, \dots, 5\}$. Each pair has a probability of $\frac{1}{36}$.

Notice the symmetry in the probabilities: $P(X=k) = P(X=-k)$ for $k \neq 0$. This property will simplify our calculations.

Step 3: Calculate the Expected Value $E[X]$.

• What we are doing: We apply the formula $E[X] = \sum x_i P(X=x_i)$ using the distribution from Step 2.
• Why we are doing this: $E[X]$ is a necessary component for calculating the variance.
• Calculations: $E[X] = \sum_{x_i=-5}^{5} x_i P(X=x_i)$ $E[X] = \frac{1}{36} \left[ (-5 \cdot 1) + (-4 \cdot 2) + (-3 \cdot 3) + (-2 \cdot 4) + (-1 \cdot 5) + (0 \cdot 6) + (1 \cdot 5) + (2 \cdot 4) + (3 \cdot 3) + (4 \cdot 2) + (5 \cdot 1) \right]$ $E[X] = \frac{1}{36} \left[ -5 - 8 - 9 - 8 - 5 + 0 + 5 + 8 + 9 + 8 + 5 \right]$ Due to the symmetry of the distribution ($P(X=k) = P(X=-k)$), each positive term cancels out its corresponding negative term: $E[X] = \frac{1}{36} \left[ (-5+5) + (-8+8) + (-9+9) + (-8+8) + (-5+5) + 0 \right] = \frac{1}{36} [0] = 0$ So, $E[X] = 0$.

Step 4: Calculate the Expected Value $E[X^2]$.

• What we are doing: We apply the formula $E[X^2] = \sum x_i^2 P(X=x_i)$.
• Why we are doing this: $E[X^2]$ is the other necessary component for calculating the variance.
• Calculations: $E[X^2] = \sum_{x_i=-5}^{5} x_i^2 P(X=x_i)$ Since $x_i^2 = (-x_i)^2$, the terms for $k$ and $-k$ will be identical. $E[X^2] = \frac{1}{36} \left[ ((-5)^2 \cdot 1) + ((-4)^2 \cdot 2) + ((-3)^2 \cdot 3) + ((-2)^2 \cdot 4) + ((-1)^2 \cdot 5) + (0^2 \cdot 6) \right.$ $\left. \qquad \qquad + (1^2 \cdot 5) + (2^2 \cdot 4) + (3^2 \cdot 3) + (4^2 \cdot 2) + (5^2 \cdot 1) \right]$ $E[X^2] = \frac{1}{36} \left[ (25 \cdot 1) + (16 \cdot 2) + (9 \cdot 3) + (4 \cdot 4) + (1 \cdot 5) + (0) \right.$ $\left. \qquad \qquad + (1 \cdot 5) + (4 \cdot 4) + (9 \cdot 3) + (16 \cdot 2) + (25 \cdot 1) \right]$ $E[X^2] = \frac{1}{36} \left[ 25 + 32 + 27 + 16 + 5 + 0 + 5 + 16 + 27 + 32 + 25 \right]$ Grouping symmetric terms: $E[X^2] = \frac{1}{36} \left[ 2(25) + 2(32) + 2(27) + 2(16) + 2(5) + 0 \right]$ $E[X^2] = \frac{1}{36} \left[ 50 + 64 + 54 + 32 + 10 \right]$ $E[X^2] = \frac{1}{36} [210]$ Simplifying the fraction by dividing numerator and denominator by their greatest common divisor, 6: $E[X^2] = \frac{210 \div 6}{36 \div 6} = \frac{35}{6}$

Step 5: Calculate the Variance $Var[X]$.

• What we are doing: We use the formula $Var[X] = E[X^2] - (E[X])^2$ with the values calculated in previous steps.
• Why we are doing this: This is the primary objective of the first part of the problem.
• Calculations: Substitute $E[X^2] = \frac{35}{6}$ and $E[X] = 0$: $Var[X] = \frac{35}{6} - (0)^2$ $Var[X] = \frac{35}{6}$ The problem states that the variance is $\frac{p}{q}$, where $p$ and $q$ are co-prime. Here, $p=35$ and $q=6$. To confirm they are co-prime: Prime factors of $35$ are $\{5, 7\}$. Prime factors of $6$ are $\{2, 3\}$. They share no common prime factors, so $35$ and $6$ are indeed co-prime.

Step 6: Find the Sum of Positive Divisors of $p$.

• What we are doing: We find the sum of positive divisors for $p=35$.
• Why we are doing this: This is the final requirement of the question.
• Calculations: The value of $p$ is $35$. First, find the prime factorization of $35$: $35 = 5^1 \cdot 7^1$. Using the sum of divisors formula $\sigma(n) = (1+p_1+\cdots+p_1^{e_1})(1+p_2+\cdots+p_2^{e_2})$: $\sigma(35) = (1+5^1)(1+7^1)$ $\sigma(35) = (1+5)(1+7)$ $\sigma(35) = (6)(8)$ $\sigma(35) = 48$ Alternatively, listing the divisors of 35: $\{1, 5, 7, 35\}$. Sum of divisors $= 1 + 5 + 7 + 35 = 48$.

3. Common Mistakes & Tips

• Leverage Symmetry: Always check if the probability distribution is symmetric. If $P(X=k) = P(X=-k)$, then $E[X]$ will be $0$, significantly simplifying calculations.
• Alternative Variance Calculation: For independent random variables $\alpha$ and $\beta$, $Var[\alpha-\beta] = Var[\alpha] + Var[\beta]$. For a single fair die roll, $E[\alpha] = 3.5$ and $E[\alpha^2] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{91}{6}$. So, $Var[\alpha] = \frac{91}{6} - (3.5)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182-147}{12} = \frac{35}{12}$. Since $\alpha$ and $\beta$ are independent and identically distributed, $Var[\alpha-\beta] = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6}$. This is a much faster method if you are familiar with these properties.
• Co-prime Check: Ensure that after calculating the variance $\frac{p}{q}$, you simplify the fraction to its lowest terms so that $p$ and $q$ are truly co-prime before proceeding to the next step.
• Arithmetic Errors: Be meticulous with additions and multiplications, especially when summing many terms for $E[X^2]$.

4. Summary

We began by defining the random variable $X = \alpha - \beta$ and constructing its probability distribution by enumerating all possible outcomes of two dice rolls. We observed the symmetry of the distribution, which allowed us to quickly determine $E[X]=0$. We then calculated $E[X^2]$ by summing $x_i^2 P(X=x_i)$ for all possible values. Using the variance formula, $Var[X] = E[X^2] - (E[X])^2$, we found the variance to be $\frac{35}{6}$. Identifying $p=35$ and $q=6$ as co-prime, the final step was to calculate the sum of the positive divisors of $p=35$, which is 48.

The final answer is $\boxed{\text{48}}$, which corresponds to option (A).