1. **Problem Statement:** We have a bag with 5 white tokens and 3 black tokens. Three tokens are drawn randomly without replacement. Let $X$ be the number of black tokens drawn. We need to find the probability distribution of $X$, then calculate the mean and variance of the linear function $g(X) = 3X + 1$, and verify the properties: (i) $E(3X+1) = 3E(X) + 1$ (ii) $\mathrm{Var}(3X+1) = 9\mathrm{Var}(X)$ 2. **Step 1: Find the probability distribution of $X$** - Total tokens: $5 + 3 = 8$ - Number of tokens drawn: 3 - $X$ can be $0, 1, 2,$ or $3$ black tokens. The probability $P(X = k)$ is given by the hypergeometric distribution: $$ P(X = k) = \frac{\binom{3}{k} \binom{5}{3-k}}{\binom{8}{3}} $$ where $k = 0,1,2,3$. Calculate each: - $P(X=0) = \frac{\binom{3}{0} \binom{5}{3}}{\binom{8}{3}} = \frac{1 \times 10}{56} = \frac{10}{56}$ - $P(X=1) = \frac{\binom{3}{1} \binom{5}{2}}{\binom{8}{3}} = \frac{3 \times 10}{56} = \frac{30}{56}$ - $P(X=2) = \frac{\binom{3}{2} \binom{5}{1}}{\binom{8}{3}} = \frac{3 \times 5}{56} = \frac{15}{56}$ - $P(X=3) = \frac{\binom{3}{3} \binom{5}{0}}{\binom{8}{3}} = \frac{1 \times 1}{56} = \frac{1}{56}$ 3. **Step 2: Calculate $E(X)$ (mean of $X$)** $$ E(X) = \sum_{k=0}^3 k P(X=k) = 0 \times \frac{10}{56} + 1 \times \frac{30}{56} + 2 \times \frac{15}{56} + 3 \times \frac{1}{56} = \frac{0 + 30 + 30 + 3}{56} = \frac{63}{56} = \frac{9}{8} = 1.125 $$ 4. **Step 3: Calculate $E(X^2)$** $$ E(X^2) = \sum_{k=0}^3 k^2 P(X=k) = 0^2 \times \frac{10}{56} + 1^2 \times \frac{30}{56} + 2^2 \times \frac{15}{56} + 3^2 \times \frac{1}{56} = \frac{0 + 30 + 60 + 9}{56} = \frac{99}{56} $$ 5. **Step 4: Calculate $\mathrm{Var}(X)$** $$ \mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \frac{99}{56} - \left(\frac{9}{8}\right)^2 = \frac{99}{56} - \frac{81}{64} $$ Find common denominator $= 448$: $$ \frac{99}{56} = \frac{99 \times 8}{448} = \frac{792}{448}, \quad \frac{81}{64} = \frac{81 \times 7}{448} = \frac{567}{448} $$ So, $$ \mathrm{Var}(X) = \frac{792}{448} - \frac{567}{448} = \frac{225}{448} \approx 0.5022 $$ 6. **Step 5: Calculate $E(g(X))$ where $g(X) = 3X + 1$** Using linearity of expectation: $$ E(g(X)) = E(3X + 1) = 3E(X) + 1 = 3 \times \frac{9}{8} + 1 = \frac{27}{8} + 1 = \frac{27}{8} + \frac{8}{8} = \frac{35}{8} = 4.375 $$ 7. **Step 6: Calculate $\mathrm{Var}(g(X))$** Using variance property for linear functions: $$ \mathrm{Var}(g(X)) = \mathrm{Var}(3X + 1) = 3^2 \mathrm{Var}(X) = 9 \times \frac{225}{448} = \frac{2025}{448} \approx 4.518 $$ 8. **Step 7: Verify properties:** (i) $E(3X+1) = 3E(X) + 1$ is shown in Step 5. (ii) $\mathrm{Var}(3X+1) = 9 \mathrm{Var}(X)$ is shown in Step 6. **Final answers:** - Probability distribution: - $P(X=0) = \frac{10}{56}$ - $P(X=1) = \frac{30}{56}$ - $P(X=2) = \frac{15}{56}$ - $P(X=3) = \frac{1}{56}$ - $E(X) = \frac{9}{8} = 1.125$ - $\mathrm{Var}(X) = \frac{225}{448} \approx 0.5022$ - $E(3X+1) = \frac{35}{8} = 4.375$ - $\mathrm{Var}(3X+1) = \frac{2025}{448} \approx 4.518$