Imagine I have set of numbers $(0,n]$. The problem I have in mind is to find the probability that the number I pick (randomly), from this set, is a power of $2$.
So I thought that I could use simply calculate the ratio of number of desired terms to total numbers in the set. Lets say there's a number $k$ such that $k$ is a power of $2$ and $k\leq n$. So, there must be $\log_{2}{k}$ other powers of $2$ that are contained in the set. I've used this reasoning to say that the probability must therefore be: $$ P=\frac{\log_{2}{n}}{n}=\frac{\ln{n}}{n\ln{2}}$$ My question is: Is this result correct? I understand if this must only be an approximation, in which case does a more accurate expression exist?
EDIT:
There's an addition I think I should mention: I could re-define the probability in terms of the floor function as: $$ P=\frac{\left \lfloor{\log_{2}{n}}\right \rfloor}{n}$$ Is this correct?
$\endgroup$ 11 Answer
$\begingroup$Almost correct.
Since $1$ is also a power of $2$, the probability is $\dfrac{\left\lfloor{\log_{2}{n}}\right\rfloor\color\red{+1}}{n}$
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