1. **Problem statement:** Given the joint probability function $$f(x,y) = \frac{1}{54}(3x + 2y - 4)$$ for $$x = 1, 2, 3$$ and $$y = 1, 2, 3$$, find the marginal probability functions of $$X$$ and $$Y$$. 2. **Formula and rules:** The marginal probability function of $$X$$ is found by summing the joint probabilities over all values of $$Y$$: $$f_X(x) = \sum_y f(x,y)$$ Similarly, the marginal probability function of $$Y$$ is found by summing over all values of $$X$$: $$f_Y(y) = \sum_x f(x,y)$$ 3. **Find marginal probability function of $$X$$:** Calculate $$f_X(x)$$ for each $$x$$: - For $$x=1$$: $$f_X(1) = \sum_{y=1}^3 \frac{1}{54}(3(1) + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (3 + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (2y - 1)$$ Calculate the sum inside: $$2(1)-1=1, \quad 2(2)-1=3, \quad 2(3)-1=5$$ Sum: $$1 + 3 + 5 = 9$$ So, $$f_X(1) = \frac{9}{54} = \frac{1}{6}$$ - For $$x=2$$: $$f_X(2) = \frac{1}{54} \sum_{y=1}^3 (3(2) + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (6 + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (2 + 2y)$$ Calculate the sum inside: $$2 + 2(1) = 4, \quad 2 + 2(2) = 6, \quad 2 + 2(3) = 8$$ Sum: $$4 + 6 + 8 = 18$$ So, $$f_X(2) = \frac{18}{54} = \frac{1}{3}$$ - For $$x=3$$: $$f_X(3) = \frac{1}{54} \sum_{y=1}^3 (3(3) + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (9 + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (5 + 2y)$$ Calculate the sum inside: $$5 + 2(1) = 7, \quad 5 + 2(2) = 9, \quad 5 + 2(3) = 11$$ Sum: $$7 + 9 + 11 = 27$$ So, $$f_X(3) = \frac{27}{54} = \frac{1}{2}$$ 4. **Find marginal probability function of $$Y$$:** Calculate $$f_Y(y)$$ for each $$y$$: - For $$y=1$$: $$f_Y(1) = \sum_{x=1}^3 \frac{1}{54}(3x + 2(1) - 4) = \frac{1}{54} \sum_{x=1}^3 (3x + 2 - 4) = \frac{1}{54} \sum_{x=1}^3 (3x - 2)$$ Calculate the sum inside: $$3(1)-2=1, \quad 3(2)-2=4, \quad 3(3)-2=7$$ Sum: $$1 + 4 + 7 = 12$$ So, $$f_Y(1) = \frac{12}{54} = \frac{2}{9}$$ - For $$y=2$$: $$f_Y(2) = \frac{1}{54} \sum_{x=1}^3 (3x + 4 - 4) = \frac{1}{54} \sum_{x=1}^3 3x = \frac{1}{54} (3 + 6 + 9) = \frac{18}{54} = \frac{1}{3}$$ - For $$y=3$$: $$f_Y(3) = \frac{1}{54} \sum_{x=1}^3 (3x + 6 - 4) = \frac{1}{54} \sum_{x=1}^3 (3x + 2)$$ Calculate the sum inside: $$3(1)+2=5, \quad 3(2)+2=8, \quad 3(3)+2=11$$ Sum: $$5 + 8 + 11 = 24$$ So, $$f_Y(3) = \frac{24}{54} = \frac{4}{9}$$ **Final answers:** - Marginal probability function of $$X$$: $$f_X(1) = \frac{1}{6}, \quad f_X(2) = \frac{1}{3}, \quad f_X(3) = \frac{1}{2}$$ - Marginal probability function of $$Y$$: $$f_Y(1) = \frac{2}{9}, \quad f_Y(2) = \frac{1}{3}, \quad f_Y(3) = \frac{4}{9}$$