We are given the following:

• The density function of $X$ is: 2x & \text{for } 0 < x < 1, \\ 0 & \text{otherwise.} \end{cases}$$
• The density function of $Y$ is uniform on the interval (1, 2), so: 1 & \text{for } 1 < y < 2, \\ 0 & \text{otherwise.} \end{cases}$$

Since $X$ and $Y$ are independent, the joint density function $f_{X,Y}(x, y)$ is the product of the individual densities:

2x \cdot 1 = 2x & \text{for } 0 < x < 1 \text{ and } 1 < y < 2, \\ 0 & \text{otherwise.} \end{cases}$$ Thus, the joint density is: $$f_{X,Y}(x, y) = \begin{cases} 2x & \text{for } 0 < x < 1 \text{ and } 1 < y < 2, \\ 0 & \text{otherwise.} \end{cases}$$ ### To calculate $$P(Y - X \geq \frac{3}{2})$$: We need to find the region where $$Y - X \geq \frac{3}{2}$$, or equivalently: $$Y \geq X + \frac{3}{2}.$$ We can now set up the integral to calculate the probability: $$P(Y - X \geq \frac{3}{2}) = \int_0^1 \int_{x + \frac{3}{2}}^2 2x \, dy \, dx.$$ First, integrate with respect to $$y$$: $$\int_{x + \frac{3}{2}}^2 2x \, dy = 2x \cdot \left( 2 - \left( x + \frac{3}{2} \right) \right) = 2x \left( \frac{1}{2} - x \right).$$ Now, integrate with respect to $$x$$ from 0 to 1: $$P(Y - X \geq \frac{3}{2}) = \int_0^1 2x \left( \frac{1}{2} - x \right) dx.$$ Simplifying the integrand: $$2x \left( \frac{1}{2} - x \right) = x - 2x^2.$$ Now, integrate: $$\int_0^1 \left( x - 2x^2 \right) dx = \left[ \frac{x^2}{2} - \frac{2x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}.$$ So, the probability is: $$P(Y - X \geq \frac{3}{2}) = \frac{1}{6}.$$ Let me know if you need further clarification! Here are some related questions to explore: 1. How does the independence of $$X$$ and $$Y$$ impact the joint distribution? 2. How do you interpret the result $$P(Y - X \geq \frac{3}{2}) = \frac{1}{6}$$? 3. What is the concept of conditional probability in the context of joint distributions? 4. How do changes in the density functions of $$X$$ and $$Y$$ affect the joint density? 5. What would happen if $$X$$ and $$Y$$ were not independent? Tip: Always double-check the limits of integration, especially when dealing with inequalities involving random variables!