A set ‘S’ contains 7 elements. A non – empty subset A of S and an element x of S are chosen at random. Then the probability that $x \in A$ is:
\[\left( A \right)\] 1/2
\[\left( B \right)\] 31/128
\[\left( C \right)\] 63/128
\[\left( D \right)\] 64/127

Hint –In this particular question use the concept that the total number of ways when an element is selected at random is the multiplication of the number of ways x can be chosen and the number of ways subset A can be chosen from the given set S respectively and to evaluate the favorable number of outcomes first find out the number of ways that x cannot be the element of A using above written technique so use this concept to reach the solution of the question.

Complete step-by-step answer:
Consider a set S having 7 elements
S = {a, b, c, d, e, f, h}
Now A is a non-empty subset of S.
I.e. $A \subseteq S$ and A $ \ne \phi $, where $\phi $ is a empty set
And x is an element of S, i.e. x belongs to S.
I.e. $x \in S$
Now we have to find the probability that $x \in A$
I.e. P ($x \in A$), where P is the symbol for probability.
Now the number of ways to select the set A from the given set S are (${2^7} - 1$)
And the number of ways to select x from S is 7 ways as there are 7 elements in set S.
So the total number of such combinations such that a non – empty subset A of S and an element x of S are chosen at random = 7(${2^7} - 1$).
So the total number of outcomes = 7(${2^7} - 1$)................... (1)
Now first find out the cases when x is not an element of A.
Now x can be chosen in 7 ways.
And in this case, a non-empty subset A such that ‘x’ is not an element of A can be chosen in (${2^6} - 1$) ways.
So the total number of ways such that $x \notin A$ are = 7(${2^6} - 1$).
So the favorable number of outcomes = 7(${2^6} - 1$).................... (2)
Now as we know that the probability (P) is the ratio of the favorable number of outcomes to the total number of outcomes so we have,
$ \Rightarrow P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$
So the probability (${P_1}$) that $x \notin A$ is = $\dfrac{{7\left( {{2^6} - 1} \right)}}{{7\left( {{2^7} - 1} \right)}}$
Now as we know that the total probability is always 1 so the probability that P ($x \in A$) is
Therefore, P ($x \in A$) = 1 – $\dfrac{{7\left( {{2^6} - 1} \right)}}{{7\left( {{2^7} - 1} \right)}}$
Now simplify this we have,
Therefore, P ($x \in A$) = $1 - \dfrac{{\left( {64 - 1} \right)}}{{\left( {128 - 1} \right)}}$
$ \Rightarrow P\left( {x \in A} \right) = 1 - \dfrac{{63}}{{127}} = \dfrac{{127 - 63}}{{127}} = \dfrac{{64}}{{127}}$
So this is the required probability.
Hence option (D) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that the probability (P) is the ratio of the favorable number of outcomes to the total number of outcomes, $P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$, and the value of the total probability is always 1, so first find out the probability that x does not belong to A then subtract this value from 1, we will get the required probability.