1. **Problem Statement:** We have two random variables: - $X$: number of heads in two coin tosses. - $Y$: result of rolling a 6-sided die once. We need to find: - (a) Probability distributions of $X$ and $Y$. - (b) Mean and variance of $X$, $Y$, $X+Y$, and $4X - 2Y$. 2. **Probability Distribution of $X$:** $X$ can be 0, 1, or 2 heads. - $P(X=0)$: both tails, probability $= \frac{1}{4}$. - $P(X=1)$: one head, one tail, probability $= \frac{2}{4} = \frac{1}{2}$. - $P(X=2)$: both heads, probability $= \frac{1}{4}$. 3. **Probability Distribution of $Y$:** $Y$ takes values 1 to 6, each with probability $\frac{1}{6}$. 4. **Mean and Variance Formulas:** - Mean $E(Z) = \sum z_i P(Z=z_i)$. - Variance $Var(Z) = E(Z^2) - [E(Z)]^2$. - For independent variables $X$ and $Y$: - $E(X+Y) = E(X) + E(Y)$. - $Var(X+Y) = Var(X) + Var(Y)$. - $Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)$. 5. **Calculate Mean and Variance of $X$:** - $E(X) = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{1}{2} = 1$. - $E(X^2) = 0^2 \times \frac{1}{4} + 1^2 \times \frac{1}{2} + 2^2 \times \frac{1}{4} = 0 + \frac{1}{2} + 1 = \frac{3}{2}$. - $Var(X) = E(X^2) - [E(X)]^2 = \frac{3}{2} - 1^2 = \frac{3}{2} - 1 = \frac{1}{2}$. 6. **Calculate Mean and Variance of $Y$:** - $E(Y) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$. - $E(Y^2) = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}$. - $Var(Y) = E(Y^2) - [E(Y)]^2 = \frac{91}{6} - (3.5)^2 = \frac{91}{6} - 12.25 = \frac{91}{6} - \frac{73.5}{6} = \frac{17.5}{6} = \frac{35}{12}$. 7. **Calculate Mean and Variance of $X + Y$:** - $E(X+Y) = E(X) + E(Y) = 1 + 3.5 = 4.5$. - $Var(X+Y) = Var(X) + Var(Y) = \frac{1}{2} + \frac{35}{12} = \frac{6}{12} + \frac{35}{12} = \frac{41}{12}$. 8. **Calculate Mean and Variance of $4X - 2Y$:** - $E(4X - 2Y) = 4E(X) - 2E(Y) = 4 \times 1 - 2 \times 3.5 = 4 - 7 = -3$. - $Var(4X - 2Y) = 4^2 Var(X) + (-2)^2 Var(Y) = 16 \times \frac{1}{2} + 4 \times \frac{35}{12} = 8 + \frac{140}{12} = 8 + \frac{35}{3} = \frac{24}{3} + \frac{35}{3} = \frac{59}{3}$. **Final answers:** - Probability distribution of $X$: $P(X=0)=\frac{1}{4}$, $P(X=1)=\frac{1}{2}$, $P(X=2)=\frac{1}{4}$. - Probability distribution of $Y$: $P(Y=k)=\frac{1}{6}$ for $k=1,2,3,4,5,6$. - $E(X)=1$, $Var(X)=\frac{1}{2}$. - $E(Y)=3.5$, $Var(Y)=\frac{35}{12}$. - $E(X+Y)=4.5$, $Var(X+Y)=\frac{41}{12}$. - $E(4X - 2Y)=-3$, $Var(4X - 2Y)=\frac{59}{3}$.