Events are:
2 is smallest or 4 is smallest or ... or 'n' is smallest number.
Lets consider the probablity that 1 is smallest no. in set S
I choose 2 with probablity 1/2.
Now, I can choose 0 more elements in (n-1)C_{0 }* $(\frac{1}{2})^{n-1}$
$(\frac{1}{2})^{n-1}$ because all elements were deselected each with probablity 1/2.
OR
I can choose 1 more elements in (n-1)C_{1}_{ }* $(\frac{1}{2})^{n-1}$
(n-1)C_{1}_{ }* $(\frac{1}{2})^{n-1}$ because I selected 1 element with prob. 1/2 and deselected others each with prob. 1/2
OR
.
.
.
I can choose n-1 more elements in (n-1)C_{1}_{ }* $(\frac{1}{2})^{n-1}$
(n-1)C_{(n-1)}_{ }* $(\frac{1}{2})^{n-1}$ because I selected (n-1) elements each with prob. 1/2
Thus, total probablity that 2 is least number
= $\frac{1}{2} * (\frac{1}{2})^{n-1} * (\binom{n-1}{0} + \binom{n-1}{1}+...+\binom{n-1}{n-1})$
=1/2
Similarly, probablity of 4 being least number = 1/4
.
.
.
Similarly, probablity of 2^{n} being least number = $\frac{1}{2^{n}}$
Now, mean
= $\sum x.P(x)$
= 2 * (1/2) + 4 * (1/4) + ... + 2^{n} * $\frac{1}{2^{n}}$
= 1 + 1 + 1 + ... + (n times)
= n