A random variable X has the following probability distribution:...

W e know $∑ {P}_{i} = 1 ⇒ 6{k}^{2} + 5k = 1$

$6{k}^{2} + 5k - 1 = 0$

$6{k}^{2} + 6k - k - 1 = 0$

\left(6k - 1\right)\left(k + 1\right) = 0 ⇒ k = - 1$(rejected) ;$k = \dfrac{1}{6}$</p><p>$P\left(X > 2\right) = k + 2k + 5{k}^{2}$</p><p>$= \dfrac{1}{6} + \dfrac{2}{6} + \dfrac{5}{36} = \dfrac{6 + 12 + 5}{36} = \dfrac{23}{36}