1. **Problem Statement:** Given the joint density function $$f(x,y) = xy^2 + \frac{x^2}{8}$$ for $$0 \leq x \leq 2$$ and $$0 \leq y \leq 1$$, find the probabilities: - $$P(X > 1)$$ - $$P\left(Y < \frac{1}{2}\right)$$ - $$P\left(X > 1 \mid Y < \frac{1}{2}\right)$$ - $$P(X + Y \leq 1)$$ 2. **Check if $$f(x,y)$$ is a valid joint density:** We verify $$\int_0^2 \int_0^1 f(x,y) \, dy \, dx = 1$$. Calculate: $$\int_0^2 \int_0^1 \left(xy^2 + \frac{x^2}{8}\right) dy \, dx = \int_0^2 \left[ x \int_0^1 y^2 dy + \frac{x^2}{8} \int_0^1 dy \right] dx$$ Evaluate inner integrals: $$\int_0^1 y^2 dy = \frac{1}{3}, \quad \int_0^1 dy = 1$$ So: $$\int_0^2 \left( x \cdot \frac{1}{3} + \frac{x^2}{8} \cdot 1 \right) dx = \int_0^2 \left( \frac{x}{3} + \frac{x^2}{8} \right) dx$$ Integrate: $$\int_0^2 \frac{x}{3} dx = \frac{1}{3} \cdot \frac{x^2}{2} \Big|_0^2 = \frac{1}{3} \cdot 2 = \frac{2}{3}$$ $$\int_0^2 \frac{x^2}{8} dx = \frac{1}{8} \cdot \frac{x^3}{3} \Big|_0^2 = \frac{1}{8} \cdot \frac{8}{3} = \frac{1}{3}$$ Sum: $$\frac{2}{3} + \frac{1}{3} = 1$$ So $$f(x,y)$$ is a valid joint density. 3. **Find $$P(X > 1)$$:** $$P(X > 1) = \int_1^2 \int_0^1 f(x,y) \, dy \, dx = \int_1^2 \int_0^1 \left(xy^2 + \frac{x^2}{8}\right) dy \, dx$$ Inner integral: $$\int_0^1 y^2 dy = \frac{1}{3}, \quad \int_0^1 dy = 1$$ So: $$\int_1^2 \left( x \cdot \frac{1}{3} + \frac{x^2}{8} \right) dx = \int_1^2 \left( \frac{x}{3} + \frac{x^2}{8} \right) dx$$ Integrate: $$\int_1^2 \frac{x}{3} dx = \frac{1}{3} \cdot \frac{x^2}{2} \Big|_1^2 = \frac{1}{6} (4 - 1) = \frac{3}{6} = \frac{1}{2}$$ $$\int_1^2 \frac{x^2}{8} dx = \frac{1}{8} \cdot \frac{x^3}{3} \Big|_1^2 = \frac{1}{24} (8 - 1) = \frac{7}{24}$$ Sum: $$P(X > 1) = \frac{1}{2} + \frac{7}{24} = \frac{12}{24} + \frac{7}{24} = \frac{19}{24}$$ 4. **Find $$P\left(Y < \frac{1}{2}\right)$$:** $$P\left(Y < \frac{1}{2}\right) = \int_0^2 \int_0^{1/2} f(x,y) \, dy \, dx = \int_0^2 \int_0^{1/2} \left(xy^2 + \frac{x^2}{8}\right) dy \, dx$$ Inner integrals: $$\int_0^{1/2} y^2 dy = \frac{(1/2)^3}{3} = \frac{1}{24}$$ $$\int_0^{1/2} dy = \frac{1}{2}$$ So: $$\int_0^2 \left( x \cdot \frac{1}{24} + \frac{x^2}{8} \cdot \frac{1}{2} \right) dx = \int_0^2 \left( \frac{x}{24} + \frac{x^2}{16} \right) dx$$ Integrate: $$\int_0^2 \frac{x}{24} dx = \frac{1}{24} \cdot \frac{x^2}{2} \Big|_0^2 = \frac{1}{24} \cdot 2 = \frac{1}{12}$$ $$\int_0^2 \frac{x^2}{16} dx = \frac{1}{16} \cdot \frac{x^3}{3} \Big|_0^2 = \frac{1}{16} \cdot \frac{8}{3} = \frac{1}{6}$$ Sum: $$P\left(Y < \frac{1}{2}\right) = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}$$ 5. **Find $$P\left(X > 1 \mid Y < \frac{1}{2}\right)$$:** By definition: $$P\left(X > 1 \mid Y < \frac{1}{2}\right) = \frac{P\left(X > 1, Y < \frac{1}{2}\right)}{P\left(Y < \frac{1}{2}\right)}$$ Calculate numerator: $$P\left(X > 1, Y < \frac{1}{2}\right) = \int_1^2 \int_0^{1/2} f(x,y) \, dy \, dx = \int_1^2 \int_0^{1/2} \left(xy^2 + \frac{x^2}{8}\right) dy \, dx$$ Inner integrals same as before: $$\int_0^{1/2} y^2 dy = \frac{1}{24}, \quad \int_0^{1/2} dy = \frac{1}{2}$$ So: $$\int_1^2 \left( x \cdot \frac{1}{24} + \frac{x^2}{8} \cdot \frac{1}{2} \right) dx = \int_1^2 \left( \frac{x}{24} + \frac{x^2}{16} \right) dx$$ Integrate: $$\int_1^2 \frac{x}{24} dx = \frac{1}{24} \cdot \frac{x^2}{2} \Big|_1^2 = \frac{1}{48} (4 - 1) = \frac{3}{48} = \frac{1}{16}$$ $$\int_1^2 \frac{x^2}{16} dx = \frac{1}{16} \cdot \frac{x^3}{3} \Big|_1^2 = \frac{1}{16} \cdot \frac{7}{3} = \frac{7}{48}$$ Sum numerator: $$\frac{1}{16} + \frac{7}{48} = \frac{3}{48} + \frac{7}{48} = \frac{10}{48} = \frac{5}{24}$$ Divide by denominator $$P\left(Y < \frac{1}{2}\right) = \frac{1}{4}$$: $$P\left(X > 1 \mid Y < \frac{1}{2}\right) = \frac{5/24}{1/4} = \frac{5}{24} \times \frac{4}{1} = \frac{20}{24} = \frac{5}{6}$$ 6. **Find $$P(X + Y \leq 1)$$:** Region: $$0 \leq x \leq 2$$, $$0 \leq y \leq 1$$, but restricted to $$x + y \leq 1$$. Rewrite: $$y \leq 1 - x$$ Limits for $$x$$: from 0 to 1 (since $$x \leq 1$$ to keep $$y \geq 0$$). So: $$P(X + Y \leq 1) = \int_0^1 \int_0^{1-x} \left(xy^2 + \frac{x^2}{8}\right) dy \, dx$$ Inner integral: $$\int_0^{1-x} y^2 dy = \frac{(1-x)^3}{3}$$ $$\int_0^{1-x} dy = 1 - x$$ So: $$\int_0^1 \left( x \cdot \frac{(1-x)^3}{3} + \frac{x^2}{8} (1-x) \right) dx = \int_0^1 \left( \frac{x(1-x)^3}{3} + \frac{x^2(1-x)}{8} \right) dx$$ Expand terms: $$\frac{x(1-x)^3}{3} = \frac{x(1 - 3x + 3x^2 - x^3)}{3} = \frac{x - 3x^2 + 3x^3 - x^4}{3}$$ $$\frac{x^2(1-x)}{8} = \frac{x^2 - x^3}{8}$$ Sum integrand: $$\frac{x - 3x^2 + 3x^3 - x^4}{3} + \frac{x^2 - x^3}{8}$$ Find common denominator 24: $$\frac{8x - 24x^2 + 24x^3 - 8x^4}{24} + \frac{3x^2 - 3x^3}{24} = \frac{8x - 24x^2 + 24x^3 - 8x^4 + 3x^2 - 3x^3}{24}$$ Combine like terms: $$8x + (-24x^2 + 3x^2) + (24x^3 - 3x^3) - 8x^4 = 8x - 21x^2 + 21x^3 - 8x^4$$ Integral: $$\int_0^1 \frac{8x - 21x^2 + 21x^3 - 8x^4}{24} dx = \frac{1}{24} \int_0^1 (8x - 21x^2 + 21x^3 - 8x^4) dx$$ Integrate term by term: $$\int_0^1 8x dx = 4$$ $$\int_0^1 -21x^2 dx = -7$$ $$\int_0^1 21x^3 dx = \frac{21}{4} = 5.25$$ $$\int_0^1 -8x^4 dx = -\frac{8}{5} = -1.6$$ Sum: $$4 - 7 + 5.25 - 1.6 = (4 - 7) + (5.25 - 1.6) = -3 + 3.65 = 0.65$$ Multiply by $$\frac{1}{24}$$: $$P(X + Y \leq 1) = \frac{0.65}{24} = \frac{13}{480}$$ **Final answers:** - $$P(X > 1) = \frac{19}{24}$$ - $$P\left(Y < \frac{1}{2}\right) = \frac{1}{4}$$ - $$P\left(X > 1 \mid Y < \frac{1}{2}\right) = \frac{5}{6}$$ - $$P(X + Y \leq 1) = \frac{13}{480}$$