To solve this, I think we need to use combinatorial reasoning.=
Consider a bit string of length ( n ). There are ( 2^n ) possible bit strings of this length because each bit can independently be either 0 or 1.
Let's denote:
- ( k ) as the number of 0's in the string
- ( n ) as the length of the bit string
This means:
- For ( k = 0 ), there's only one possibility, which is all bits being 1. So, there's 1 string.
- For ( k = 1 ), there are ( n ) possibilities because there are ( n ) positions where the single 0 can be placed in the string.
- For ( k = 1 ), there are ( \binom{n}{1} ) possibilities because there are ( n ) positions where the single 0 can be placed in the string.
- For ( k = 2 ), there are
\( \binom{n}{2} \)possibilities because we're choosing 2 positions out of ( n ) for the 0's to occupy. -For ( k = 3 ), there are ( \binom{n}{3} ) possibilities, and so on, until ( k = \lfloor \frac{n}{2} \rfloor )
The asnwer: total number of bit strings with more 0's than 1's is the sum of the possibilities for each ( k ) from 0 to ( \lfloor \frac{n}{2} \rfloor ).
Is this right?