Let $\zeta(s):=\sum_{n=1}^\infty \frac{1}{n^s}$.For fixed $s>1$ let the random variable $X$ with $\mathbb P(X=n)=\frac{1}{\zeta(s)n^s}$ and let $A_m:=\{{X \textrm{ is divisiable by m}}\}$ I want to calculate $\mathbb P(\frac{X}{n}=k | A_n), k >0, n\in \mathbb N$. My attempt:
$$\mathbb P(\frac{X}{n}=k|A_n)=\mathbb P(X=kn|A_n)=\frac{\mathbb P((X=kn) \cap A_n)}{\mathbb P(A_n)}=\frac{\mathbb P(X=kn)\mathbb P(A_n)}{\mathbb P(A_n)}=\mathbb P(X=kn)$$ but this is nonsense. Help is much appreciated!
We have that $$ (X = kn) \in A_{\,n} \quad \Rightarrow \quad (X = kn) \cap A_{\,n} = (X = kn) $$ therefore
$$\mathbb P(\frac{X}{n}=k|A_n)=\mathbb P(X=kn|A_n)=\frac{\mathbb P((X=kn) \cap A_n)}{\mathbb P(A_n)}=\frac{\mathbb P(X=kn)}{\mathbb P(A_n)}$$
Now $$ A_{\,n} = \left\{ {X = jn\;\,\left| {\;j \in \mathbb N} \right.} \right\} $$ and $$ \mathbb P\left( {A_{\,n} } \right) = \sum\limits_{1\, \le \;j} {\mathbb P\left( {X = jn} \right)} = \sum\limits_{1\, \le \;j} {{1 \over {\zeta (s)\left( {jn} \right)^{\,s} }}} $$ So $$ \eqalign{ & {{\mathbb P\left( {X = kn} \right)} \over {\mathbb P\left( {A_{\,n} } \right)}} = {1 \over {\zeta (s)\left( {kn} \right)^{\,s} \sum\limits_{1\, \le \;j} {{1 \over {\zeta (s)\left( {jn} \right)^{\,s} }}} }} = \cr & = {1 \over {k^{\,s} \sum\limits_{1\, \le \;j} {{1 \over {j^{\,s} }}} }} = {1 \over {\zeta (s)k^{\,s} }} = \mathbb P(X=k)\cr} $$