I am learning about combinatorial and bit strings.
I decided to use combinatorial reasoning and wanted to see if my logic made sense.
The question: How many bit strings of length n contain more 0’s than 1’s?
Note: the length is unknown
Let there be a bit string of length ( n )
with (k) as the number of 0s. We need to ensure
$ k > (n - k)$ as we want to show more 0s and 1s.
This can be simplify into $( 2k > n )$ or $\frac{n}{2} < k$
This can also be represented as a combination formula $\binom{n}{k}$ with the total number of bit strings being $\binom{n}{k}$ for all values of ( k ) satisfying k > $\frac{n}{2}$
We can represent it as
$\sum_{k = \lceil \frac{n}{2} \rceil}^{n} \binom{n}{k}$
where the sum of the binomial coefficients for choosing k positions out of n where k is greater than $\frac{n}{2}$