1. The possible sums \(x\) that can be obtained when a pair of dice is rolled range from 2 to 12.
2. Calculate the probability function \( P(X = x) \):
P(X = x) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
P(X = x) = \begin{cases} \frac{x-1}{36} & x = 2,3,4,5,6,7 \\\frac{13-x}{36} & x = 8,9,10,11,12 \\0 & \text{otherwise}\end{cases}
3. Calculate the mathematical expectation \( E(X) \):
E(X) = \sum_{x=2}^{12} x \cdot P(X = x)
E(X) = \sum_{x=2}^{7} x \cdot \frac{x-1}{36} + \sum_{x=8}^{12} x \cdot \frac{13-x}{36}
4. Simplify the summation to find \( E(X) \):
E(X) = \sum_{x=2}^{7} \frac{x(x-1)}{36} + \sum_{x=8}^{12} \frac{x(13-x)}{36}
E(X) = 7
5. Calculate the variance \( \text{Var}(X) \):
\text{Var}(X) = E(X^2) - [E(X)]^2
6. Calculate \( E(X^2) \):
E(X^2) = \sum_{x=2}^{12} x^2 \cdot P(X = x)
7. Simplify the summation to find \( E(X^2) \) and then \( \text{Var}(X) \):
E(X^2) = \sum_{x=2}^{7} \frac{x^2(x-1)}{36} + \sum_{x=8}^{12} \frac{x^2(13-x)}{36}
\text{Var}(X) = E(X^2) - [E(X)]^2
\text{Var}(X) = \frac{35}{6}
Therefore, the probability function, mathematical expectation, and variance are:
P(X = x) = \begin{cases} \frac{x-1}{36} & x = 2,3,4,5,6,7 \\\frac{13-x}{36} & x = 8,9,10,11,12 \\0 & \text{otherwise}\end{cases}
E(X) = 7
\text{Var}(X) = \frac{35}{6}