Consider the probability distribution $\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline…

The probability distribution requires $\sum \mathrm{P}(\mathrm{X}=x) = 1$. With $\mathrm{P}(1) = \mathrm{K}$, $\mathrm{P}(2) = 2\mathrm{K}$, $\mathrm{P}(3) = \mathrm{K}^2$, $\mathrm{P}(4) = 2\mathrm{K}$, and $\mathrm{P}(5) = 5\mathrm{K}^2$, the sum is $5\mathrm{K} + 6\mathrm{K}^2 = 1$.

Solving $6\mathrm{K}^2 + 5\mathrm{K} - 1 = 0$ yields $\mathrm{K} = \frac{-5 \pm 7}{12}$. Only the positive solution $\mathrm{K} = \frac{1}{6}$ is valid since probabilities must be non-negative.

The probability $\mathrm{P}(\mathrm{X} > 2)$ is the sum $\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) = \mathrm{K}^2 + 2\mathrm{K} + 5\mathrm{K}^2 = 6\mathrm{K}^2 + 2\mathrm{K}$.

Substituting $\mathrm{K} = \frac{1}{6}$ gives $6\left(\frac{1}{6}\right)^2 + 2\left(\frac{1}{6}\right) = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$.

The value $\frac{1}{2}$ corresponds to option C.