Geometric distribution over the natural numbers

Question

I’m thinking of the geometric distribution over $\mathbb{N}$, i.e. $P(\{i\})=\frac{1}{2^i}$. I know $P$ is defined on the power set of $\mathbb{N}$. But is it true that for every rational number $0\le\frac{m}{n}\le 1$, there is an event $A$ such that $P(A)=\frac{m}{n}$? I can’t think of a counterexample but I can’t prove it as well.

Answer 1

Indeed, every number $\alpha$ between $0,1$ has a unique binary decimal expansion (with the usual remark about ending in an infinite string of $1's$), so just let $A_{\alpha}$ denote the event "you select a natural number $n$ for which the $n^{th}$ digit of the binary expansion of $\alpha$ is $1$."

Note that $\alpha$ need not be rational.

For instance $$\frac 37=0.011011011011011\dots_2$$

So the associated event would be selecting $n\equiv 0,2\pmod 3$

For rational $\alpha$, the expansion is eventually periodic, so you essentially come down to congruences.

And $$\pi -3 =0.00100100001111110110101010001\dots_2$$

So the associated event would be selecting $n\in \{3, 6,11,12,13,14,15,16,18, \cdots\}$.