1. **Problem Statement:** We have a random variable $X$ defined as the sum of the faces when a pair of dice is thrown. We need to find the expected value $E(X)$ and variance $Var(X)$ of $X$. 2. **Possible values of $X$:** When two dice are thrown, the sum $X$ can take values from 2 to 12. 3. **Probability distribution:** The probability $P(X=x)$ for each sum $x$ is given by the number of ways to get $x$ divided by total outcomes (36). Number of ways for each sum: - 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 4. **Expected value formula:** $$E(X) = \sum_x x P(X=x)$$ Calculate: $$E(X) = \frac{1}{36}(2\times1 + 3\times2 + 4\times3 + 5\times4 + 6\times5 + 7\times6 + 8\times5 + 9\times4 + 10\times3 + 11\times2 + 12\times1)$$ Calculate the sum inside: $$= \frac{1}{36}(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)$$ $$= \frac{252}{36} = 7$$ So, $E(X) = 7$. 5. **Variance formula:** $$Var(X) = E(X^2) - [E(X)]^2$$ Calculate $E(X^2)$: $$E(X^2) = \sum_x x^2 P(X=x) = \frac{1}{36}(2^2\times1 + 3^2\times2 + 4^2\times3 + 5^2\times4 + 6^2\times5 + 7^2\times6 + 8^2\times5 + 9^2\times4 + 10^2\times3 + 11^2\times2 + 12^2\times1)$$ Calculate the sum inside: $$= \frac{1}{36}(4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144)$$ $$= \frac{1974}{36} = 54.8333...$$ 6. **Calculate variance:** $$Var(X) = 54.8333 - 7^2 = 54.8333 - 49 = 5.8333$$ 7. **Final answers:** - Expected value $E(X) = 7$ - Variance $Var(X) = \frac{35}{6} \approx 5.8333$