The random variable takes the values $1,2,3,..................,m$. If $P(X = n) = \dfrac{1}{m}$ to each $n$, then the value of $X$ is:
A. $\dfrac{{(m + 1)(2m + 1)}}{6}$
B. $\dfrac{{{m^2} - 1}}{{12}}$$n$
C. $\dfrac{{m + 1}}{2}$
D. $\dfrac{{{m^2} + 1}}{{12}}$

Hint: According to the question given in the question we have to determine the value of $X$ when the random variable takes the values$1,2,3,..................,m$. If $P(X = n) = \dfrac{1}{m}$ to each so, first of all we have to understand about the variance which is explained below:
Variance: Variance is a measurement of the spread between numbers in the given data set, so it measures how far each number in the set is from the mean and therefore from every other number in the set.
Now, we have to find the value of $E(x)$ with the formula as mentioned below:

Formula used:
$ E(x) = \sum\limits_{i = 1}^{i = n} {{x_i}{p_i}} .......................(a)$
Sum of n natural numbers $ = \dfrac{{n(n + 1)}}{2}...........................(b)$
Now, as given in the question we have to substitute the value of $P(X = n) = \dfrac{1}{m}$ as in the formula (a) above as mentioned in the question.
Now, after obtaining the value of $E(x)$with the help of formula (a) we have to determine the variance with the help of the formula as given below:
$E({x^2}) = \sum\limits_{i = 1}^{i = n} {{x_i}^2{p_i}} ......................(c)$
Sum of squares of all the natural numbers $ = \dfrac{{n(n + 1)(2n + 1)}}{6}......................(d)$
Hence, after obtaining the value of $E({x^2})$ with the help of the formula (b) we have to find the value of (X).

Complete step-by-step solution:
Step 1: First of all we have to obtain the value of $E(x)$ with the help of the formula (a) as mentioned in the solution hint. Hence, on substituting all the values in formula (a),
$E(x) = \left( {1 \times \dfrac{1}{m} + 2 \times \dfrac{2}{m} + 3 \times \dfrac{3}{m}.........m \times \dfrac{1}{m}} \right)$
On solving the expression obtained just above,
$E(x) = (1 + 2 + 3 + ............m) \times \dfrac{1}{m}..........................(1)$
Step 2: Now, to solve the expression (1) as obtained in the solution step 1 have to use the formula (b) as mentioned in the solution hint.
$
   \Rightarrow E(x) = \dfrac{{m(m + 1)}}{2} \times \dfrac{1}{m} \\
   \Rightarrow E(x) = \dfrac{{m + 1}}{2}.................(2) \\
 $
Step 3: Now, we have to find the value of $E({x^2})$ with the help of the formula (c) as mentioned in the solution hint. Hence, substituting all the values in the formula (c),
$ \Rightarrow E({x^2}) = \left( {{1^2} \times \dfrac{1}{m} + {2^2} \times \dfrac{2}{m} + {3^2} \times \dfrac{3}{m} + ....................{m^2} \times \dfrac{1}{m}} \right)$
On solving the expression obtained just above,
$ \Rightarrow E({x^2}) = ({1^2} + {2^2} + {3^2} + ......................{m^2}) \times \dfrac{1}{m}......................(2)$
Step 4: Now, to solve the expression as obtained in step 3 we have to use the formula (d) as mentioned in the solution hint. Hence,
$
   \Rightarrow E({x^2}) = \dfrac{{m(m + 1)(2m + 1)}}{6} \times \dfrac{1}{m} \\
   \Rightarrow E({x^2}) = \dfrac{{(m + 1)(2m + 1)}}{6}.....................(3)
 $
Step 5: Now, to find the value of X we have to use the formula (e) as mentioned in the solution hint so, we have to substitute the obtained expression (1) and expression (3) in the formula (e),
$ \Rightarrow E({x^2}) - {\left[ {E(x)} \right]^2} = \dfrac{{(m + 1)(2m + 1)}}{6} - \dfrac{{{{(m + 1)}^2}}}{4}$
Now, to find the value of X we have to take the L.C.M of the expression obtained just above,
$
   \Rightarrow X = \dfrac{{(m + 1)(4m + 2 - 3m - 3)}}{{12}} \\
   \Rightarrow X = \dfrac{{(m + 1)(m - 1)}}{{12}} \\
   \Rightarrow X = \dfrac{{{m^2} - 1}}{{12}}
 $
Final solution: Hence, with the help of the formula (a), (b), (c), and (d) we have obtained the value of X which is $ = \dfrac{{{m^2} - 1}}{{12}}$

Therefore option (B) is correct.

Note: Variance is a measurement of the spread between a given number of the data set that is, it measures how far each number in the set is from the mean and hence, from every other number in the set.
We can easily find the sum of all the natural number as (1, 2, 3, 4, ………………..n) with the help of $\dfrac{{n(n + 1)}}{2}$ and same as we can obtain the sum of all the squares of natural numbers as $({1^2},{2^2},{3^2},{4^2},............{n^2})$ with the help of $\dfrac{{n(n + 1)(2n + 1)}}{6}$.