I am not sure how to solve this, but this is my best guess. Since we want to know how many 12 bit strings have more 1's than 0's, start with 5 since it is one less than 12/2=6.
Then we proceed with:
$${12 \choose 5} + {12 \choose 4} + {12 \choose 3} + {12 \choose 2}+ {12 \choose 1}+ {12 \choose 0}=1586$$
Is my reasoning correct? I am not sure if we should use the choose function here, but I believe we do.
Best Answer
Your expression is correct. Here is another way that exploits the symmetry.
Let $a$ be the number of strings with more $1$'s than $0$'s, let $b$ be the number with more $0$'s than $1$'s, and let $c$ be the number with equal numbers of $0$'s and $1$'s.
Then $a=b$, $a+b+c=2^{12}$, and $c=\binom{12}{6}$. So $2a=2^{12}-\binom{12}{6}$ and therefore $$a=2^{11}-\frac{1}{2}\binom{12}{6}.$$