1. **Problem 9(a): Find the marginal probability function of X** given joint probability function $$f(x,y) = \frac{1}{54}(3x + 2y - 4)$$ for $$x=1,2,3$$ and $$y=1,2,3$$. 2. The marginal probability function of X is found by summing over all values of Y: $$f_X(x) = \sum_{y=1}^3 f(x,y) = \sum_{y=1}^3 \frac{1}{54}(3x + 2y - 4)$$ 3. Calculate for each $$x$$: - For $$x=1$$: $$f_X(1) = \frac{1}{54} \sum_{y=1}^3 (3(1) + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (3 + 2y - 4) = \frac{1}{54} \sum_{y=1}^3 (-1 + 2y)$$ Calculate terms: $$y=1: -1 + 2(1) = 1$$ $$y=2: -1 + 2(2) = 3$$ $$y=3: -1 + 2(3) = 5$$ 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 terms: $$y=1: 2 + 2(1) = 4$$ $$y=2: 2 + 2(2) = 6$$ $$y=3: 2 + 2(3) = 8$$ 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 terms: $$y=1: 5 + 2(1) = 7$$ $$y=2: 5 + 2(2) = 9$$ $$y=3: 5 + 2(3) = 11$$ Sum: $$7 + 9 + 11 = 27$$ So, $$f_X(3) = \frac{27}{54} = \frac{1}{2}$$ 4. **Answer for 9(a):** $$f_X(1) = \frac{1}{6}, \quad f_X(2) = \frac{1}{3}, \quad f_X(3) = \frac{1}{2}$$ 5. **Problem 9(b): Find the marginal probability function of Y** by summing over all values of X: $$f_Y(y) = \sum_{x=1}^3 f(x,y) = \sum_{x=1}^3 \frac{1}{54}(3x + 2y - 4)$$ 6. Calculate for each $$y$$: - For $$y=1$$: $$f_Y(1) = \frac{1}{54} \sum_{x=1}^3 (3x + 2(1) - 4) = \frac{1}{54} \sum_{x=1}^3 (3x + 2 - 4) = \frac{1}{54} \sum_{x=1}^3 (3x - 2)$$ Calculate terms: $$x=1: 3(1) - 2 = 1$$ $$x=2: 3(2) - 2 = 4$$ $$x=3: 3(3) - 2 = 7$$ 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 terms: $$x=1: 3(1) + 2 = 5$$ $$x=2: 3(2) + 2 = 8$$ $$x=3: 3(3) + 2 = 11$$ Sum: $$5 + 8 + 11 = 24$$ So, $$f_Y(3) = \frac{24}{54} = \frac{4}{9}$$ 7. **Answer for 9(b):** $$f_Y(1) = \frac{2}{9}, \quad f_Y(2) = \frac{1}{3}, \quad f_Y(3) = \frac{4}{9}$$ --- 8. **Problem 10(a): Construct the joint probability function table for X and Y** where X = number of dice showing 1, Y = number of dice showing 4, 5, or 6, rolling two balanced dice. 9. Possible values for X and Y are 0, 1, or 2 because there are two dice. 10. Total outcomes = 36. Each die is independent. 11. Calculate probabilities for each (X,Y): - For each die, probability of 1 is $$\frac{1}{6}$$. - Probability of 4,5,6 is $$\frac{3}{6} = \frac{1}{2}$$. - Probability of other numbers (2,3) is $$\frac{2}{6} = \frac{1}{3}$$. 12. We find joint probabilities by counting outcomes matching (X,Y): - (X=0,Y=0): neither die is 1 nor 4,5,6, so both dice show 2 or 3. Probability = $$\left(\frac{2}{6}\right)^2 = \frac{4}{36} = \frac{1}{9}$$ - (X=0,Y=1): no dice show 1, exactly one die shows 4,5,6. Number of ways: choose which die shows 4,5,6 (2 ways), other die shows 2 or 3. Probability = $$2 \times \frac{1}{2} \times \frac{1}{3} = \frac{2}{6} = \frac{1}{3}$$ - (X=0,Y=2): no dice show 1, both dice show 4,5,6. Probability = $$\left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ - (X=1,Y=0): one die shows 1, other die shows 2 or 3. Probability = $$2 \times \frac{1}{6} \times \frac{1}{3} = \frac{2}{18} = \frac{1}{9}$$ - (X=1,Y=1): one die shows 1, one die shows 4,5,6. Probability = $$2 \times \frac{1}{6} \times \frac{1}{2} = \frac{2}{12} = \frac{1}{6}$$ - (X=1,Y=2): impossible because only two dice total. Probability = 0 - (X=2,Y=0): both dice show 1. Probability = $$\left(\frac{1}{6}\right)^2 = \frac{1}{36}$$ - (X=2,Y=1): impossible. Probability = 0 - (X=2,Y=2): impossible. Probability = 0 13. Joint probability table: | X \ Y | 0 | 1 | 2 | |-------|---|---|---| | 0 | 1/9 | 1/3 | 1/4 | | 1 | 1/9 | 1/6 | 0 | | 2 | 1/36 | 0 | 0 | 14. **Problem 10(b): Find marginal probability function of X** by summing over Y: - $$f_X(0) = \frac{1}{9} + \frac{1}{3} + \frac{1}{4} = \frac{4}{36} + \frac{12}{36} + \frac{9}{36} = \frac{25}{36}$$ - $$f_X(1) = \frac{1}{9} + \frac{1}{6} + 0 = \frac{4}{36} + \frac{6}{36} = \frac{10}{36} = \frac{5}{18}$$ - $$f_X(2) = \frac{1}{36} + 0 + 0 = \frac{1}{36}$$ 15. **Problem 10(c): Find marginal probability function of Y** by summing over X: - $$f_Y(0) = \frac{1}{9} + \frac{1}{9} + \frac{1}{36} = \frac{4}{36} + \frac{4}{36} + \frac{1}{36} = \frac{9}{36} = \frac{1}{4}$$ - $$f_Y(1) = \frac{1}{3} + \frac{1}{6} + 0 = \frac{12}{36} + \frac{6}{36} = \frac{18}{36} = \frac{1}{2}$$ - $$f_Y(2) = \frac{1}{4} + 0 + 0 = \frac{9}{36} = \frac{1}{4}$$ **Final answers:** - 9(a): $$f_X(1) = \frac{1}{6}, f_X(2) = \frac{1}{3}, f_X(3) = \frac{1}{2}$$ - 9(b): $$f_Y(1) = \frac{2}{9}, f_Y(2) = \frac{1}{3}, f_Y(3) = \frac{4}{9}$$ - 10(a): Joint probability table as above. - 10(b): $$f_X(0) = \frac{25}{36}, f_X(1) = \frac{5}{18}, f_X(2) = \frac{1}{36}$$ - 10(c): $$f_Y(0) = \frac{1}{4}, f_Y(1) = \frac{1}{2}, f_Y(2) = \frac{1}{4}$$