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: (a) The marginal probability function of $X$. (b) The marginal probability function of $Y$. 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 $X$: $$f_Y(y) = \sum_x f(x,y)$$ - Since $f(x,y)$ is a probability function, the sum over all $x$ and $y$ must be 1. 3. **Find $f_X(x)$:** Calculate 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 terms: - $y=1$: $2(1)-1=1$ - $y=2$: $2(2)-1=3$ - $y=3$: $2(3)-1=5$ 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 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. **Find $f_Y(y)$:** Calculate 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 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, $$f_Y(3) = \frac{24}{54} = \frac{4}{9}$$ 5. **Summary:** - 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 $Y$: $$f_Y(1) = \frac{2}{9}, \quad f_Y(2) = \frac{1}{3}, \quad f_Y(3) = \frac{4}{9}$$