1. **Problem statement:** Given the joint probability density function (pdf) of random variables $X$ and $Y$: $$f_{X,Y}(x,y) = \frac{2}{3}(x + xy), \quad 0 < x < 1, 0 < y < 2; \quad 0 \text{ otherwise}$$ Find: (a) Marginal pdf of $X$. (b) Marginal pdf of $Y$. (c) Conditional pdf of $Y$ given $X=\frac{1}{3}$. (d) Probability $P\left(Y \leq 1 \mid X=\frac{1}{3}\right)$. 2. **Marginal pdf of $X$:** The marginal pdf of $X$ is found by integrating the joint pdf over all $y$: $$f_X(x) = \int_0^2 f_{X,Y}(x,y) \, dy = \int_0^2 \frac{2}{3}(x + xy) \, dy$$ 3. **Calculate $f_X(x)$:** $$f_X(x) = \frac{2}{3} \int_0^2 (x + xy) \, dy = \frac{2}{3} \left[ x y + x \frac{y^2}{2} \right]_0^2 = \frac{2}{3} \left( 2x + x \cdot 2 \right) = \frac{2}{3} \cdot 4x = \frac{8}{3} x$$ for $0 < x < 1$. 4. **Marginal pdf of $Y$:** Similarly, integrate over $x$: $$f_Y(y) = \int_0^1 f_{X,Y}(x,y) \, dx = \int_0^1 \frac{2}{3}(x + xy) \, dx = \frac{2}{3} \int_0^1 x(1 + y) \, dx$$ 5. **Calculate $f_Y(y)$:** $$f_Y(y) = \frac{2}{3} (1 + y) \int_0^1 x \, dx = \frac{2}{3} (1 + y) \cdot \frac{1}{2} = \frac{1}{3} (1 + y)$$ for $0 < y < 2$. 6. **Conditional pdf of $Y$ given $X=\frac{1}{3}$:** By definition: $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)}$$ Substitute $x=\frac{1}{3}$: $$f_{Y|X}\left(y \middle| \frac{1}{3}\right) = \frac{\frac{2}{3} \left( \frac{1}{3} + \frac{1}{3} y \right)}{\frac{8}{3} \cdot \frac{1}{3}} = \frac{\frac{2}{3} \cdot \frac{1}{3} (1 + y)}{\frac{8}{9}} = \frac{\frac{2}{9} (1 + y)}{\frac{8}{9}} = \frac{2}{9} (1 + y) \cdot \frac{9}{8} = \frac{1}{4} (1 + y)$$ for $0 < y < 2$. 7. **Calculate $P\left(Y \leq 1 \mid X=\frac{1}{3}\right)$:** $$P\left(Y \leq 1 \mid X=\frac{1}{3}\right) = \int_0^1 f_{Y|X}\left(y \middle| \frac{1}{3}\right) \, dy = \int_0^1 \frac{1}{4} (1 + y) \, dy = \frac{1}{4} \left[ y + \frac{y^2}{2} \right]_0^1 = \frac{1}{4} \left(1 + \frac{1}{2}\right) = \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8}$$ **Final answers:** - (a) $f_X(x) = \frac{8}{3} x, \quad 0 < x < 1$ - (b) $f_Y(y) = \frac{1}{3} (1 + y), \quad 0 < y < 2$ - (c) $f_{Y|X}\left(y \middle| \frac{1}{3}\right) = \frac{1}{4} (1 + y), \quad 0 < y < 2$ - (d) $P\left(Y \leq 1 \mid X=\frac{1}{3}\right) = \frac{3}{8}$