A set $\mathrm{S}$ contains 7 elements. A non-empty subset $A$ of $S$ and an element $x$ of $S$ are chosen…
A set $\mathrm{S}$ contains 7 elements. A non-empty subset $A$ of $S$ and an element $x$ of $S$ are chosen at random. Then the probability that $\mathrm{x} \in \mathrm{A}$ is:
$\frac{1}{2}$
$\frac{64}{127}$
$\frac{63}{128}$
$\frac{31}{128}$
Solution
Let $\mathrm{S}=\left\{x_1, x_2, x_3, x_4, x_5, x_6, x_7\right\}$
Let the chosen element be $\mathrm{x}_{\mathrm{i}}$.
Total number of subsets of $\mathrm{S}=2^7=128$
No. of non-empty subsets of S $=128-1$ $=127$
We need to find number of those subsets that contains $x_i$.
\begin{array}{|l|l|l|l|l|l|l|}
\hline 2 & 2 & 2 & 2 & 1 & 2 & 2 \\
\hline
\end{array}
$x_1 x_2$-—- $x_i=-x_7$
For those subsets containing $x_i$, each element has 2 choices.
i.e., (included or not included) in subset, However as the subset must contain $x_i$, $x_i$ has only one choice. (included one)
So, total no. of subsets containing $x_i=2 \times 2 \times 2 \times 2 \times 1 \times 2 \times 2=64$
Required prob
$
=\frac{\text { No. of subsets containing } x_i}{\text { Total no. of non-empty subsets }}
$
$
=\frac{64}{127}
$