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