With $n$ bits we can represent $2^n$ distinct numbers. In both $1's$ complement as well as sign magnitude representations we have $2$ zeroes, and the rest $2^n -2$ numbers are used to represent equal number of positive and negative numbers.
In $2's$ complement representation there is a single zero. The rest $2^n-1$ numbers are split to $2^{n-1}-1$ positive numbers and $2^{n-1}$ negative numbers.
So, in none of the given representations we have more positive numbers than negative numbers.
Correct Answer: D.