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On set $\Omega = \{ 1,2,\dots,n\}$, $n \in \mathbb{N}$ and $\sigma$-algebra, which is its power set, we denote $P$ is probability measure, where $P(\{ k \}) = \frac{1}{n}$ for all $k = 1,2,\dots,n$. Let $\mathbf{P}$ is set of prime numbers. We denote $\phi(n)$ is number of all distinct numbers of $n$. We define events $A_p := \{p,2p,3p,\dots \} \cap \{1,2,\dots, n \}$ where p is prime.

How to prove that $A_p$ are independent, where p is prime divisor of n? And then i need to prove $\frac{\phi(n)}{n} = \prod_{p\in \mathbf{P}, p|n} (1-\frac{1}{p}).$

Thank you for any help.

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$A_p$'s are not independent.

Counter-example:

Let $\Omega = \{ 1,2,3\}$, $A_2 = \{2\}$, $A_3 = \{ 3 \}$. So $$0 = P(\{\phi\}) = P(A_2,A_3) \neq P(A_2)*P(A_3) = \frac{1}{3}*\frac{1}{3}. $$

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  • $\begingroup$ Is the same if $p$ is all prime divisors of number $n$? $\endgroup$ – user10 Feb 20 '17 at 15:38
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    $\begingroup$ if $p$ is all prime divisors of number $n$ then the counter examples fails. $\endgroup$ – rookie Feb 20 '17 at 15:49
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    $\begingroup$ How to prove independence in that case? $\endgroup$ – user10 Feb 20 '17 at 16:10

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