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

Then, $$P(X > 2)$$ is equal to:

For a discrete random variable the very first property we use is: “The sum of all the probabilities is equal to 1.” Stated mathematically, if the possible values of $$X$$ are $$x_1,x_2,\dots ,x_n$$ with respective probabilities $$p_1,p_2,\dots ,p_n,$$ then

$$\displaystyle \sum_{i=1}^{n} p_i \;=\;1.$$

Here the values and probabilities are given as

$$\begin{aligned} X &: \;1 \quad 2 \quad 3 \quad 4 \quad 5,\\[2mm] P(X) &: \;k^{2}\quad 2k \quad k \quad 2k \quad 5k^{2}. \end{aligned}$$

Applying the above property, we write

$$k^{2}+2k+k+2k+5k^{2}=1.$$

Now we collect like terms:

$$\bigl(k^{2}+5k^{2}\bigr)+\bigl(2k+k+2k\bigr)=1,$$

$$6k^{2}+5k=1.$$

Next we transpose the 1 to the left side so that every term lies on the same side of the equality:

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

This is a quadratic equation in $$k$$ of the standard form $$ax^{2}+bx+c=0.$$ We now invoke the quadratic-formula, which states

$$x=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a},\quad\text{for}\;ax^{2}+bx+c=0.$$

With $$a=6,\;b=5,\;c=-1$$ we substitute:

$$k=\frac{-5\pm\sqrt{\,5^{2}-4\cdot6\cdot(-1)\,}}{2\cdot6}.$$

Inside the square root we simplify step by step:

$$5^{2}=25,\quad -4\cdot6\cdot(-1)=24,$$

so

$$\sqrt{\,25+24\,}=\sqrt{49}=7.$$

Hence

$$k=\frac{-5\pm7}{12}.$$

This yields two numerical possibilities:

$$k=\frac{-5+7}{12}=\frac{2}{12}=\frac16,\qquad k=\frac{-5-7}{12}=\frac{-12}{12}=-1.$$

Because probability values must be non-negative, we discard the negative root. Therefore

$$k=\frac16.$$

We now proceed to the requested probability $$P(X>2).$$ The condition $$X>2$$ corresponds to the outcomes $$X=3,4,5.$$ Using the definition of probability for mutually exclusive events, we add their individual probabilities:

$$P(X>2)=P(X=3)+P(X=4)+P(X=5).$$

Substituting the expressions given in the table, we get

$$P(X>2)=k+2k+5k^{2}.$$

First we combine like terms:

$$k+2k=3k,$$

so

$$P(X>2)=3k+5k^{2}.$$

Now we replace $$k$$ by the value already obtained, $$k=\dfrac16$$:

$$P(X>2)=3\left(\frac16\right)+5\left(\frac16\right)^{2}.$$

We simplify each term separately. For the linear term,

$$3\left(\frac16\right)=\frac36=\frac12.$$

For the quadratic term, first square the denominator: $$\left(\frac16\right)^{2}=\frac1{36}.$$ Then multiply by 5:

$$5\left(\frac1{36}\right)=\frac5{36}.$$

Adding the two fractions with a common denominator of 36, we have

$$\frac12+\frac5{36}=\frac{18}{36}+\frac5{36}=\frac{23}{36}.$$

Thus

$$P(X>2)=\frac{23}{36}.$$

Consulting the options given, the fraction $$\dfrac{23}{36}$$ corresponds to Option D.

Hence, the correct answer is Option D.