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Q.
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:
Probability
Solution:
Let $S=\left\{x_{1},\, x_{2},\, x_{3}, \,x_{4},\,x_{5},\, x_{6},\,x_{7}\right\}$
Let the chosen element be $x_{i}$
Total number of subsets of $S=2^{7} =128$
No. of non-empty subsets of $S=128-1=127$
We need to find number of those subsets that contains $x_{i}$.
$x_1\, x_2 ------- x_i ---- x_7$
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
$x_{i}=2 \times 2 \times 2 \times 2 \times 1 \times 2 \times 2=64$
Req. prob. $=\frac{\text { No. of subsets containing } x_{i}}{\text { Total no. of non-empty subsets }}=\frac{64}{127}$
