Question

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 :

• A. $\frac{1}{2}$
• B. $\frac{31}{128}$
• C. $\frac{63}{128}$
• D. $\frac{64}{127}$

Answer 1

Answer: (D)

$\frac{64}{127}$

$S=\{a,a_2,a_3...........a_7\}$
$S\subseteq S$ $x\in S$ $S\ne \varphi$
S can be chosen in $(2^7-1)$ ways
x can be chosen in 7 ways
$\therefore$ Total combinations = 7 $(2^7-1)$

First find the cases where 'x' is not an element of 'A'

'x' can be chosen in $7$ ways.

In this case, a non-empty subset such that 'x' is not an element can be chosen in $2^6-1$ ways.

Hence, number of ways of choosing 'x' and 'A' such that 'x' is not in 'S' is $7\times (2^6-1)$
$\therefore$ Needed probability $=1-\left[\frac{7(2^6-1)}{7(2^7-1)}\right]=\frac{64}{127}$