1. **Problem Statement:** We have a joint density function: $$f(x,y) = \frac{6 - x - y}{8} \quad \text{for } 0 \leq x \leq 2, 2 \leq y \leq 4$$ and zero otherwise. We want to calculate the probability: $$P\left(X \leq \frac{2}{3}, Y \leq \frac{5}{2}\right)$$ 2. **Understanding the Limits:** - The function is defined only for $x$ between 0 and 2, and $y$ between 2 and 4. - The event $X \leq \frac{2}{3}$ means $x$ goes from 0 up to $\frac{2}{3}$. - The event $Y \leq \frac{5}{2}$ means $y$ goes from 2 up to $\frac{5}{2}$ (since $y$ cannot be less than 2 in the domain). 3. **Setting up the Integral:** The probability is the double integral of $f(x,y)$ over the region: $$0 \leq x \leq \frac{2}{3}, \quad 2 \leq y \leq \frac{5}{2}$$ So, $$P = \int_{y=2}^{5/2} \int_{x=0}^{2/3} \frac{6 - x - y}{8} \, dx \, dy$$ 4. **Explanation:** - The inner integral integrates with respect to $x$ from 0 to $\frac{2}{3}$. - The outer integral integrates with respect to $y$ from 2 to $\frac{5}{2}$. This matches the domain restrictions and the event conditions. 5. **Summary:** The limits for $x$ are from 0 to $\frac{2}{3}$ because $X \leq \frac{2}{3}$ and $x$ must be within the domain $[0,2]$. The limits for $y$ are from 2 to $\frac{5}{2}$ because $Y \leq \frac{5}{2}$ and $y$ must be within the domain $[2,4]$. Hence, the integral limits are: $$\int_{2}^{5/2} \int_{0}^{2/3} \frac{6 - x - y}{8} \, dx \, dy$$