1. **Problem statement:** Given the joint probability density function (pdf) of $X$ and $Y$: $$p(x,y) = \begin{cases} k(8 - x - y), & 0 < x < 3, 0 < y < 4 \\ 0, & \text{otherwise} \end{cases}$$ Find the constant $k$ and the conditional expectation $E[X \mid Y=3]$. 2. **Find $k$ using the normalization condition:** The total probability must be 1: $$\int_0^4 \int_0^3 k(8 - x - y) \, dx \, dy = 1$$ 3. **Calculate the inner integral:** $$\int_0^3 (8 - x - y) \, dx = \int_0^3 (8 - y - x) \, dx = \left[(8 - y)x - \frac{x^2}{2}\right]_0^3 = (8 - y)3 - \frac{9}{2} = 24 - 3y - 4.5 = 19.5 - 3y$$ 4. **Calculate the outer integral:** $$\int_0^4 k(19.5 - 3y) \, dy = k \left[19.5y - \frac{3y^2}{2}\right]_0^4 = k(19.5 \times 4 - \frac{3 \times 16}{2}) = k(78 - 24) = 54k$$ 5. **Set equal to 1 and solve for $k$:** $$54k = 1 \implies k = \frac{1}{54}$$ 6. **Find the conditional pdf of $X$ given $Y=3$:** $$p_{X|Y}(x|3) = \frac{p(x,3)}{p_Y(3)} = \frac{k(8 - x - 3)}{p_Y(3)} = \frac{k(5 - x)}{p_Y(3)}$$ where $0 < x < 3$. 7. **Find the marginal pdf of $Y$ at $y=3$:** $$p_Y(3) = \int_0^3 p(x,3) \, dx = \int_0^3 k(8 - x - 3) \, dx = k \int_0^3 (5 - x) \, dx = k \left[5x - \frac{x^2}{2}\right]_0^3 = k(15 - 4.5) = 10.5k = 10.5 \times \frac{1}{54} = \frac{10.5}{54} = \frac{7}{36}$$ 8. **Calculate the conditional expectation:** $$E[X|Y=3] = \int_0^3 x p_{X|Y}(x|3) \, dx = \frac{1}{p_Y(3)} \int_0^3 x k (5 - x) \, dx = \frac{k}{p_Y(3)} \int_0^3 (5x - x^2) \, dx$$ 9. **Evaluate the integral:** $$\int_0^3 (5x - x^2) \, dx = \left[\frac{5x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{5 \times 9}{2} - \frac{27}{3} = \frac{45}{2} - 9 = 22.5 - 9 = 13.5$$ 10. **Substitute values:** $$E[X|Y=3] = \frac{k}{p_Y(3)} \times 13.5 = \frac{\frac{1}{54}}{\frac{7}{36}} \times 13.5 = \frac{1}{54} \times \frac{36}{7} \times 13.5 = \frac{36 \times 13.5}{54 \times 7} = \frac{486}{378} = \frac{9}{7}$$ **Final answers:** $$k = \frac{1}{54}, \quad E[X|Y=3] = \frac{9}{7}$$ **Matching with options:** This corresponds to option d) in the multiple-choice list.