If a random variable X has the following probability distribution values of X.

Then, P( X ≥ 6)=

$\frac{19}{100}$

Since the sum of the probabilities in a probability distribution is always unity.

$∴ P(X=0)+P(X=1)+.....+P(X=7)=1$

$⇒ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 +k =1 $

$⇒ 10k^2 + 9k - 1=0 $

$⇒ (10k -1) (k+1)=0$

$⇒ 10k - 1= 0 $     $[∵k ≥ 0 ∴ k + 1 ≠ 0]$

$⇒ k = \frac{1}{10}$

Now, 

$P(X ≥ 6) =P(X= 6) + P(X = 7)$

$= 2k^2 + 7k^2 + k = 9k^2 + k =\frac{19}{100}$   $[∵k =\frac{1}{10}]$