1. **Problem Statement:** We have a joint probability mass function (p.m.f) $p(x,y)$ for the number of calls $X$ and trips booked $Y$. Given $p(x,y)$ values, we need to find: (i) The marginal p.m.f of $X$ and $E(X)$. (ii) The conditional p.m.f of $Y$ given $X$. (iii) The conditional p.m.f of $X$ given $Y$. 2. **Recall formulas:** - Marginal p.m.f of $X$: $p_X(x) = \sum_y p(x,y)$. - Expected value: $E(X) = \sum_x x p_X(x)$. - Conditional p.m.f of $Y$ given $X=x$: $p_{Y|X}(y|x) = \frac{p(x,y)}{p_X(x)}$. - Conditional p.m.f of $X$ given $Y=y$: $p_{X|Y}(x|y) = \frac{p(x,y)}{p_Y(y)}$ where $p_Y(y) = \sum_x p(x,y)$. --- 3. **Calculate marginal p.m.f of $X$:** - $p_X(0) = p(0,0) = 0.04$ - $p_X(1) = p(1,0) + p(1,1) = 0.08 + 0.06 = 0.14$ - $p_X(2) = p(2,0) + p(2,1) + p(2,2) = 0.12 + 0.20 + 0.12 = 0.44$ - $p_X(3) = p(3,0) + p(3,1) + p(3,2) + p(3,3) = 0.10 + 0.16 + 0.10 + 0.02 = 0.38$ Check sum: $0.04 + 0.14 + 0.44 + 0.38 = 1.00$ (valid p.m.f) 4. **Calculate $E(X)$:** $$E(X) = 0 \times 0.04 + 1 \times 0.14 + 2 \times 0.44 + 3 \times 0.38 = 0 + 0.14 + 0.88 + 1.14 = 2.16$$ --- 5. **Calculate conditional p.m.f of $Y$ given $X=x$:** For each $x$, divide $p(x,y)$ by $p_X(x)$: - For $X=0$: $p_{Y|X}(0|0) = \frac{0.04}{0.04} = 1$ - For $X=1$: - $p_{Y|X}(0|1) = \frac{0.08}{0.14} \approx 0.571$ - $p_{Y|X}(1|1) = \frac{0.06}{0.14} \approx 0.429$ - For $X=2$: - $p_{Y|X}(0|2) = \frac{0.12}{0.44} \approx 0.273$ - $p_{Y|X}(1|2) = \frac{0.20}{0.44} \approx 0.455$ - $p_{Y|X}(2|2) = \frac{0.12}{0.44} \approx 0.273$ - For $X=3$: - $p_{Y|X}(0|3) = \frac{0.10}{0.38} \approx 0.263$ - $p_{Y|X}(1|3) = \frac{0.16}{0.38} \approx 0.421$ - $p_{Y|X}(2|3) = \frac{0.10}{0.38} \approx 0.263$ - $p_{Y|X}(3|3) = \frac{0.02}{0.38} \approx 0.053$ --- 6. **Calculate marginal p.m.f of $Y$:** Sum over $x$: - $p_Y(0) = 0.04 + 0.08 + 0.12 + 0.10 = 0.34$ - $p_Y(1) = 0.06 + 0.20 + 0.16 = 0.42$ - $p_Y(2) = 0.12 + 0.10 = 0.22$ - $p_Y(3) = 0.02$ 7. **Calculate conditional p.m.f of $X$ given $Y=y$:** - For $Y=0$: - $p_{X|Y}(0|0) = \frac{0.04}{0.34} \approx 0.118$ - $p_{X|Y}(1|0) = \frac{0.08}{0.34} \approx 0.235$ - $p_{X|Y}(2|0) = \frac{0.12}{0.34} \approx 0.353$ - $p_{X|Y}(3|0) = \frac{0.10}{0.34} \approx 0.294$ - For $Y=1$: - $p_{X|Y}(1|1) = \frac{0.06}{0.42} \approx 0.143$ - $p_{X|Y}(2|1) = \frac{0.20}{0.42} \approx 0.476$ - $p_{X|Y}(3|1) = \frac{0.16}{0.42} \approx 0.381$ - For $Y=2$: - $p_{X|Y}(2|2) = \frac{0.12}{0.22} \approx 0.545$ - $p_{X|Y}(3|2) = \frac{0.10}{0.22} \approx 0.455$ - For $Y=3$: - $p_{X|Y}(3|3) = \frac{0.02}{0.02} = 1$ --- **Final answers:** (i) Marginal p.m.f of $X$: $$p_X(0)=0.04, p_X(1)=0.14, p_X(2)=0.44, p_X(3)=0.38$$ Expected value: $$E(X) = 2.16$$ (ii) Conditional p.m.f of $Y$ given $X$: - $X=0$: $p_{Y|X}(0|0)=1$ - $X=1$: $p_{Y|X}(0|1)=0.571, p_{Y|X}(1|1)=0.429$ - $X=2$: $p_{Y|X}(0|2)=0.273, p_{Y|X}(1|2)=0.455, p_{Y|X}(2|2)=0.273$ - $X=3$: $p_{Y|X}(0|3)=0.263, p_{Y|X}(1|3)=0.421, p_{Y|X}(2|3)=0.263, p_{Y|X}(3|3)=0.053$ (iii) Conditional p.m.f of $X$ given $Y$: - $Y=0$: $p_{X|Y}(0|0)=0.118, p_{X|Y}(1|0)=0.235, p_{X|Y}(2|0)=0.353, p_{X|Y}(3|0)=0.294$ - $Y=1$: $p_{X|Y}(1|1)=0.143, p_{X|Y}(2|1)=0.476, p_{X|Y}(3|1)=0.381$ - $Y=2$: $p_{X|Y}(2|2)=0.545, p_{X|Y}(3|2)=0.455$ - $Y=3$: $p_{X|Y}(3|3)=1$