1. **Problem Statement:** (a) Given joint pdf $f_{X,Y}(x,y) = \frac{15}{2}(2 - x - y)$ for $0 < x < 1$, $0 < y < 1$, find the conditional density of $X$ given $Y = y$. 2. **Formula and Explanation:** The conditional density of $X$ given $Y = y$ is defined as: $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$ where $f_Y(y)$ is the marginal density of $Y$. 3. **Find Marginal Density $f_Y(y)$:** $$f_Y(y) = \int_0^1 f_{X,Y}(x,y) \, dx = \int_0^1 \frac{15}{2}(2 - x - y) \, dx$$ Calculate the integral: $$= \frac{15}{2} \int_0^1 (2 - y - x) \, dx = \frac{15}{2} \left[ (2 - y)x - \frac{x^2}{2} \right]_0^1 = \frac{15}{2} \left( 2 - y - \frac{1}{2} \right) = \frac{15}{2} \left( \frac{3}{2} - y \right) = \frac{15}{2} \times \frac{3 - 2y}{2} = \frac{15(3 - 2y)}{4}$$ 4. **Write Conditional Density:** $$f_{X|Y}(x|y) = \frac{\frac{15}{2}(2 - x - y)}{\frac{15(3 - 2y)}{4}} = \frac{\frac{15}{2}(2 - x - y)}{\frac{15(3 - 2y)}{4}} = \frac{(2 - x - y)}{(3 - 2y)} \times 2$$ Simplify: $$f_{X|Y}(x|y) = \frac{2(2 - x - y)}{3 - 2y}$$ for $0 < x < 1$ and $0 < y < 1$. 5. **Interpretation:** This conditional density tells us the distribution of $X$ when $Y$ is fixed at $y$. **Final answer:** $$\boxed{f_{X|Y}(x|y) = \frac{2(2 - x - y)}{3 - 2y}, \quad 0 < x < 1, 0 < y < 1}$$