1. **Problem statement:** Find the expected value and variance of the random variable $X$ defined as the sum of faces when a pair of dice is thrown. 2. **Formula and rules:** - The expected value (mean) of a discrete random variable $X$ is given by $$E(X) = \sum x_i p(x_i)$$ where $x_i$ are the possible values and $p(x_i)$ their probabilities. - The variance is $$V(X) = E(X^2) - [E(X)]^2$$ where $$E(X^2) = \sum x_i^2 p(x_i)$$. 3. **Possible values of $X$:** When two dice are thrown, the sum $X$ can be from 2 to 12. 4. **Probabilities:** The number of ways to get each sum is: - 2: 1 way - 3: 2 ways - 4: 3 ways - 5: 4 ways - 6: 5 ways - 7: 6 ways - 8: 5 ways - 9: 4 ways - 10: 3 ways - 11: 2 ways - 12: 1 way Total outcomes = 36. So, $$p(x) = \frac{\text{number of ways}}{36}$$. 5. **Calculate $E(X)$:** $$E(X) = \sum_{x=2}^{12} x p(x) = \frac{1}{36}(2\times1 + 3\times2 + 4\times3 + 5\times4 + 6\times5 + 7\times6 + 8\times5 + 9\times4 + 10\times3 + 11\times2 + 12\times1)$$ Calculate the sum inside: $$= \frac{1}{36}(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)$$ $$= \frac{252}{36} = 7$$ 6. **Calculate $E(X^2)$:** $$E(X^2) = \sum x^2 p(x) = \frac{1}{36}(2^2\times1 + 3^2\times2 + 4^2\times3 + 5^2\times4 + 6^2\times5 + 7^2\times6 + 8^2\times5 + 9^2\times4 + 10^2\times3 + 11^2\times2 + 12^2\times1)$$ Calculate the sum inside: $$= \frac{1}{36}(4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144)$$ $$= \frac{1974}{36} = 54.8333$$ 7. **Calculate variance:** $$V(X) = E(X^2) - [E(X)]^2 = 54.8333 - 7^2 = 54.8333 - 49 = 5.8333$$ **Final answers:** - Expected value $E(X) = 7$ - Variance $V(X) = 5.8333$