1. **Problem Statement:** We have a discrete random variable $X$ representing the number of heads when tossing three fair coins. The probability distribution is given by: | $X$ | 0 | 1 | 2 | 3 | |-----|---|---|---|---| | $f(x)$ | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ | We need to find: (a) $E(X)$, $E(X^2)$, and $E[X(X-1)]$ (b) $\mathrm{Var}(X)$ (c) $\mathrm{Var}(X^2)$ --- 2. **Formulas and Rules:** - Expected value: $E(g(X)) = \sum_x g(x) f(x)$ where $g(x)$ is a function of $X$. - Variance: $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$. - For $E[X(X-1)]$, treat $g(X) = X(X-1)$. --- 3. **Calculate $E(X)$:** $$E(X) = \sum_x x f(x) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8}$$ $$= 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = 1.5$$ --- 4. **Calculate $E(X^2)$:** $$E(X^2) = \sum_x x^2 f(x) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{8}$$ $$= 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$$ --- 5. **Calculate $E[X(X-1)]$:** $$E[X(X-1)] = \sum_x x(x-1) f(x)$$ Calculate each term: - For $x=0$: $0 \cdot (0-1) = 0$ - For $x=1$: $1 \cdot 0 = 0$ - For $x=2$: $2 \cdot 1 = 2$ - For $x=3$: $3 \cdot 2 = 6$ So, $$E[X(X-1)] = 0 \cdot \frac{1}{8} + 0 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 6 \cdot \frac{1}{8} = 0 + 0 + \frac{6}{8} + \frac{6}{8} = \frac{12}{8} = 1.5$$ --- 6. **Calculate $\mathrm{Var}(X)$:** $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75$$ --- 7. **Calculate $\mathrm{Var}(X^2)$:** First find $E(X^4)$: $$E(X^4) = \sum_x x^4 f(x) = 0^4 \cdot \frac{1}{8} + 1^4 \cdot \frac{3}{8} + 2^4 \cdot \frac{3}{8} + 3^4 \cdot \frac{1}{8}$$ $$= 0 + \frac{3}{8} + \frac{48}{8} + \frac{81}{8} = \frac{132}{8} = 16.5$$ Then, $$\mathrm{Var}(X^2) = E(X^4) - [E(X^2)]^2 = 16.5 - 3^2 = 16.5 - 9 = 7.5$$ --- **Final answers:** - $E(X) = 1.5$ - $E(X^2) = 3$ - $E[X(X-1)] = 1.5$ - $\mathrm{Var}(X) = 0.75$ - $\mathrm{Var}(X^2) = 7.5$