For an arbitrary string sampled from $\{0,1\}^n$, what would be the probability that the string has a Kolmogorov Complexity $\geq$ length of the string, $n$.
In other words, how many strings are there in the above set that satisfy the definition of random, as defined in Kolmogorov Complexity.
A string $x$ is $c$-compressible if $K(x) \leq |x| + c$. If $x$ is not compressible by $1$, $x$ is said to be incompressible (or random, if you will).
There are $2^n$ bit strings of length $n$, and there are $\sum_{i=0}^{n-1} 2^i = 2^n-1$ descriptions that are of length less than $n$. Because each description describes at most one string, there is at least one incompressible bit string for each length $n$.
Also, at least $2^n-2^{n-c+1}+1$ strings of length $n$ are incompressible by $c$. Once again, at most $2^{n-c+1}-1$ strings of length $n$ are $c$-compressible, because we have at most that many descriptions of length at most $n-c$. Then the remaining $2^n-(2^{n-c+1}-1)$ strings are incompressible by $c$.
