# Given Information - \(X\) and \(Y\) are discrete random variables. - Joint probability mass function: \(P(X = x, Y = y) = k(x + y)\), where \(x = , 1, 2\) and \(y = , 1, 2\). - \(P(X = x, Y = y) = \) otherwise. # What to Find 1. The value of \(k\). 2. Marginal probability mass functions of \(X\) and \(Y\). 3. Whether \(X\) and \(Y\) are independent. 4. \(P(X + Y \leq 2)\). # Definition or Concept Used - **Joint PMF**: The probability that \(X = x\) and \(Y = y\) together. - **Marginal PMF**: Probability of one variable regardless of the other, obtained by summing the joint PMF over the other variable. - **Independence**: \(X\) and \(Y\) are independent if \(P(X=x, Y=y) = P(X=x)P(Y=y)\) for all \(x, y\). - **Probability Calculation**: For any event, sum the joint probabilities over all outcomes in that event. # Step by Step Solution **Finding the value of \(k\):** The total probability must sum to 1. Sum over all \(x, y\) values: \[ \sum_{x=}^2 \sum_{y=}^2 k(x+y) = 1 \] Calculate each \(x+y\) for all combinations (\(3 \times 3\) grid): \[ \begin{align*} x=: &\ y=,1,2 \implies +, +1, +2 = ,1,2 \\ x=1: &\ y=,1,2 \implies 1+, 1+1, 1+2 = 1,2,3 \\ x=2: &\ y=,1,2 \implies 2+, 2+1, 2+2 = 2,3,4 \\ \end{align*} \] Sum: \(+1+2+1+2+3+2+3+4 = 18\) So, \[ k (+1+2+1+2+3+2+3+4) = k \times 18 = 1 \implies k = \frac{1}{18} \] **Marginal PMF of \(X\):** \[ P(X = x) = \sum_{y=}^2 P(X = x, Y = y) = \frac{1}{18} \sum_{y=}^2 (x + y) \] Calculate for each \(x\): - \(x=\): \(\frac{1}{18}(+ + +1 + +2) = \frac{1}{18}(+1+2) = \frac{3}{18} = \frac{1}{6}\) - \(x=1\): \(\frac{1}{18}(1+ + 1+1 + 1+2) = \frac{1}{18}(1+2+3) = \frac{6}{18} = \frac{1}{3}\) - \(x=2\): \(\frac{1}{18}(2+ + 2+1 + 2+2) = \frac{1}{18}(2+3+4) = \frac{9}{18} = \frac{1}{2}\) So, \[ P(X=) = \frac{1}{6},\quad P(X=1) = \frac{1}{3},\quad P(X=2) = \frac{1}{2} \] **Marginal PMF of \(Y\):** \[ P(Y = y) = \sum_{x=}^2 P(X = x, Y = y) = \frac{1}{18} \sum_{x=}^2 (x + y) \] Calculate for each \(y\): - \(y=\): \(\frac{1}{18}(+ + 1+ + 2+) = \frac{1}{18}(+1+2) = \frac{3}{18} = \frac{1}{6}\) - \(y=1\): \(\frac{1}{18}(+1 + 1+1 + 2+1) = \frac{1}{18}(1+2+3) = \frac{6}{18} = \frac{1}{3}\) - \(y=2\): \(\frac{1}{18}(+2 + 1+2 + 2+2) = \frac{1}{18}(2+3+4) = \frac{9}{18} = \frac{1}{2}\) So, \[ P(Y=) = \frac{1}{6},\quad P(Y=1) = \frac{1}{3},\quad P(Y=2) = \frac{1}{2} \] **Are \(X\) and \(Y\) independent?** If independent, \(P(X=x, Y=y) = P(X=x)P(Y=y)\) for all \(x, y\). Check for \(x=, y=\): \(P(X=, Y=) = \frac{1}{18}(+)=\), \(P(X=)P(Y=) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \neq \) So, \(X\) and \(Y\) are **not independent**. **Compute \(P(X + Y \leq 2)\):** Possible \((x, y)\) pairs: - \((,): \) - \((,1): 1\) - \((,2): 2\) - \((1,): 1\) - \((1,1): 2\) - \((2,): 2\) Sum their probabilities: \[ \begin{align*} P(,) &= \frac{1}{18} \times = \\ P(,1) &= \frac{1}{18} \times 1 = \frac{1}{18} \\ P(,2) &= \frac{1}{18} \times 2 = \frac{2}{18} \\ P(1,) &= \frac{1}{18} \times 1 = \frac{1}{18} \\ P(1,1) &= \frac{1}{18} \times 2 = \frac{2}{18} \\ P(2,) &= \frac{1}{18} \times 2 = \frac{2}{18} \\ \end{align*} \] Total: \(\frac{1+2+1+2+2}{18} = \frac{8}{18} = \frac{4}{9}\) # Final Answers - \(k = \frac{1}{18}\) - Marginal PMF: - \(P(X=) = \frac{1}{6},\ P(X=1) = \frac{1}{3},\ P(X=2) = \frac{1}{2}\) - \(P(Y=) = \frac{1}{6},\ P(Y=1) = \frac{1}{3},\ P(Y=2) = \frac{1}{2}\) - \(X\) and \(Y\) are **not independent**. - \(P(X + Y \leq 2) = \frac{4}{9}\)

# Given Information - \(X\) and \(Y\) are discrete random variables. - Joint probability mass function: \(P(X = z, Y = y) = k(z + y)\) for \(z = 0, 1, 2\) and \(y = 0, 1, 2\). - \(P(X = z, Y = y) = 0\) otherwise. # What to Find 1. The value of \(k\). 2. Marginal probability mass functions of \(X\) and \(Y\). 3. Whether \(X\) and \(Y\) are independent. 4. Compute \(P(X + Y \leq 2)\). # Definitions or Concepts Used - **Joint PMF**: Probability that \(X = z\) and \(Y = y\) occur simultaneously. - **Marginal PMF**: Probability of one variable irrespective of the other, calculated by summing the joint PMF over the other variable. - **Independence**: \(X\) and \(Y\) are independent if \(P(X=z, Y=y) = P(X=z)P(Y=y)\) for all values of \(z\) and \(y\). - **Probability Calculation**: The total probability must sum to 1 across all possible outcomes. # Step-by-Step Solution **Finding the Value of \(k\):** The total probability must sum to 1. We compute: \[ \sum_{z=0}^2 \sum_{y=0}^2 P(X=z, Y=y) = 1 \] Calculating the joint probabilities for all combinations: \[ \begin{align*} P(X=0, Y=0) &= k(0+0) = 0 \\ P(X=0, Y=1) &= k(0+1) = k \\ P(X=0, Y=2) &= k(0+2) = 2k \\ P(X=1, Y=0) &= k(1+0) = k \\ P(X=1, Y=1) &= k(1+1) = 2k \\ P(X=1, Y=2) &= k(1+2) = 3k \\ P(X=2, Y=0) &= k(2+0) = 2k \\ P(X=2, Y=1) &= k(2+1) = 3k \\ P(X=2, Y=2) &= k(2+2) = 4k \\ \end{align*} \] Summing these: \[ 0 + k + 2k + k + 2k + 3k + 2k + 3k + 4k = 18k = 1 \implies k = \frac{1}{18} \] **Marginal PMF of \(X\):** \[ P(X=z) = \sum_{y=0}^2 P(X=z, Y=y) = k \sum_{y=0}^2 (z + y) \] Calculating for each \(z\): - For \(z=0\): \[ P(X=0) = k(0 + 0 + 1 + 2) = k(3) = \frac{1}{18} \cdot 3 = \frac{1}{6} \] - For \(z=1\): \[ P(X=1) = k(1 + 0 + 1 + 2) = k(4) = \frac{1}{18} \cdot 4 = \frac{2}{9} \] - For \(z=2\): \[ P(X=2) = k(2 + 0 + 1 + 2) = k(5) = \frac{1}{18} \cdot 5 = \frac{5}{18} \] **Marginal PMF of \(Y\):** \[ P(Y=y) = \sum_{z=0}^2 P(X=z, Y=y) = k \sum_{z=0}^2 (z + y) \] Calculating for each \(y\): - For \(y=0\): \[ P(Y=0) = k(0 + 0 + 1 + 2) = k(3) = \frac{1}{6} \] - For \(y=1\): \[ P(Y=1) = k(0 + 1 + 1 + 2) = k(4) = \frac{2}{9} \] - For \(y=2\): \[ P(Y=2) = k(0 + 2 + 1 + 2) = k(5) = \frac{5}{18} \] **Checking Independence of \(X\) and \(Y\):** We check if \(P(X=z, Y=y) = P(X=z)P(Y=y)\): - For \(z=0, y=0\): \[ P(X=0, Y=0) = 0 \quad \text{and} \quad P(X=0)P(Y=0) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \neq 0 \] Thus, \(X\) and \(Y\) are **not independent**. **Computing \(P(X + Y \leq 2)\):** Identify pairs \((X, Y)\) such that \(X + Y \leq 2\): - \((0, 0)\) → \(P(X=0, Y=0) = 0\) - \((0, 1)\) → \(P(X=0, Y=1) = k = \frac{1}{18}\) - \((0, 2)\) → \(P(X=0, Y=2) = 2k = \frac{2}{18} = \frac{1}{9}\) - \((1, 0)\) → \(P(X=1, Y=0) = k = \frac{1}{18}\) - \((1, 1)\) → \(P(X=1, Y=1) = 2k = \frac{2}{18} = \frac{1}{9}\) Sum their probabilities: \[ P(X + Y \leq 2) = P(0, 0) + P(0, 1) + P(0, 2) + P(1, 0) + P(1, 1) = 0 + \frac{1}{18} + \frac{1}{9} + \frac{1}{18} + \frac{1}{9} \] Calculating: \[ = 0 + \frac{1}{18} + \frac{2}{18} + \frac{1}{18} + \frac{2}{18} = \frac{6}{18} = \frac{1}{3} \] # Final Answers - \(k = \frac{1}{18}\) - Marginal PMF: - \(P(X=0) = \frac{1}{6},\ P(X=1) = \frac{2}{9},\ P(X=2) = \frac{5}{18}\) - \(P(Y=0) = \frac{1}{6},\ P(Y=1) = \frac{2}{9},\ P(Y=2) = \frac{5}{18}\) - \(X\) and \(Y\) are **not independent**. - \(P(X + Y \leq 2) = \frac{1}{3}\)