Solution :
Given the Joint Probability distribution is :
Y|X 0 1 2 3 P(y)
0 k 0.03 0.03 0.03 k+0.09
1 0.01 0.02 0.04 0.04 0.11
2 0.01 0.01 0.05 0.72 0.79
P(x) k+0.02 0.06 0.12 0.79 k+0.99
Since we know,
[tex]$\sum \sum P(x_i,y_j) = 1$[/tex]
⇒ k + 0.99 = 1
⇒ k = 1 - 0.99
⇒ k = 0.01
Therefore, the value of k is 0.01