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The random variables X and Y have a discrete joint distribution with joint probability function:
P(X = x, Y = y) = c(x + 2y); x = 0, 1, 2; y = 0, 1, 2
Determine the value of c.
A
B
C
1
D
Explanation:
The correct answer is B) .
Explanation:
For a valid joint probability distribution, the sum of all probabilities must equal 1:
Substituting the given probability function:
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: 27`c = 1c = \frac{1}{27}$$
Therefore, the correct value of c is .