What is the probability of that a number in the set of natural numbers is divisible by $3$ but not by $9$? 
What is the probability of that a number in the set of natural numbers is divisible by $3$ but not by $9$?

Any ideas about this question? How should I approach this problem?
 A: HINT:
$$1,2,\color{red}3,4,5,\color{red}6,7,8,9\bigg|10,11,\color{red}{12},13,14,\color{red}{15},16,17,18\bigg|19,20,\color{red}{21},22,23,\color{red}{24},25,26,27\bigg|28,\cdots$$
Each section contains $9$ numbers and only $2$ of them satisfy the conditions...
A: Probability that a natural number is divisible by $3$ is $\frac 13$.
Probability that a natural number that is known to be divisible by $3$ is divisible by $9$ is $\frac 13$.
Hence probability that a natural number that is known to be divisible by $3$ is not divisible by $9$ is $\frac 23$.
Therefore probability that a natural number is divisible by $3$ but not by $9$ is $\frac 23 \cdot \frac 13 = \frac 29$.
A: This question somehow reveals a problem concerning probability distributions on countably infinite sets. 
The point is that there cannot exist equiprobability for all elements of $\mathbb N$. 
If we would state that $p$ denotes the probability of the occurence of number $n\in\mathbb N$ then both possibilities $p=0$ and $p>0$ appear to lead to something that is not true:


*

*if $p=0$ then $\mathsf P(X\in\mathbb N)=\sum_{n\in\mathbb N}\mathsf P(X=n)=\sum_{n\in\mathbb N}p=0\neq1$.

*if $p>0$ then $\mathsf P(X\in\mathbb N)=\sum_{n\in\mathbb N}\mathsf P(X=n)=\sum_{n\in\mathbb N}p=+\infty\neq1$.


Actually we are forced to choose for $P(X=n)=p_n$ with $p_n\geq0$ and $\sum_{n\in\mathbb N}p_n=1$ and consequently not all $p_n$ have the same value.
That leads to answer:$$\sum_{n=1}^{\infty}p_{9n-6}+\sum_{n=1}^{\infty}p_{9n-3}$$
depending on a choice that has been made.
So not really an answer that satisfies.
