Answer: $\frac{1}{27}$

The correct answer is B) $\frac{1}{27}$. **Explanation:** For a valid joint probability distribution, the sum of all probabilities must equal 1: $$\sum_{x=0}^{2} \sum_{y=0}^{2} P(X = x, Y = y) = 1$$ Substituting the given probability function: $$\sum_{x=0}^{2} \sum_{y=0}^{2} c(x + 2y) = 1$$ Let's compute the sum step by step: For each x value: - When x = 0: c(0 + 2y) = 2cy for y = 0,1,2 Sum for x=0: c(0+0) + c(0+2) + c(0+4) = 0c + 2c + 4c = 6c - When x = 1: c(1 + 2y) for y = 0,1,2 Sum for x=1: c(1+0) + c(1+2) + c(1+4) = 1c + 3c + 5c = 9c - When x = 2: c(2 + 2y) for y = 0,1,2 Sum for x=2: c(2+0) + c(2+2) + c(2+4) = 2c + 4c + 6c = 12c Total sum: 6c + 9c + 12c = 27c Setting this equal to 1: $$27c = 1$$ $$c = \frac{1}{27}$$ Therefore, the correct value of c is $\frac{1}{27}$.

Author: Nikitesh Somanthe