1. **Problem statement:** Given the joint pdf $$f_{X,Y}(x,y) = \frac{15}{2}(2 - x - y)$$ for $$0 < x < 1$$ and $$0 < y < 1$$, find the conditional density of $$X$$ given $$Y = y$$. 2. **Recall the formula for conditional density:** $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$ where $$f_Y(y)$$ is the marginal density of $$Y$$. 3. **Find the marginal density $$f_Y(y)$$:** Integrate the joint pdf over $$x$$: $$f_Y(y) = \int_0^1 f_{X,Y}(x,y) \, dx = \int_0^1 \frac{15}{2}(2 - x - y) \, dx$$ 4. **Calculate the integral:** $$f_Y(y) = \frac{15}{2} \int_0^1 (2 - x - y) \, dx = \frac{15}{2} \left[ (2 - y)x - \frac{x^2}{2} \right]_0^1 = \frac{15}{2} \left( (2 - y) \cdot 1 - \frac{1}{2} \right) = \frac{15}{2} \left( \frac{3}{2} - y \right) = \frac{15}{2} \left( \frac{3 - 2y}{2} \right) = \frac{15}{4} (3 - 2y)$$ 5. **Write the conditional density:** $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{\frac{15}{2}(2 - x - y)}{\frac{15}{4}(3 - 2y)} = \frac{\frac{15}{2}(2 - x - y)}{\frac{15}{4}(3 - 2y)} = \frac{(2 - x - y)}{(3 - 2y)} \times 2$$ 6. **Simplify:** $$f_{X|Y}(x|y) = \frac{2(2 - x - y)}{3 - 2y}$$ 7. **Domain:** Since $$0 < x < 1$$ and $$0 < y < 1$$, the conditional density is valid for $$x$$ in $$0 < x < 1$$ given $$y$$ in $$0 < y < 1$$. **Final answer:** $$f_{X|Y}(x|y) = \frac{2(2 - x - y)}{3 - 2y}$$ for $$0 < x < 1$$ and $$0 < y < 1$$.