If (Ω,Pr) is the probability space where |Ω| is prime, and Pr is the uniform probability distribution on Ω, how would you show that any two non-trivial events A and B cannot be independent, (they must be either positively or negatively correlated).
I know that 2 events are independent if Pr(A∩B) = Pr(A)⋅Pr(B), positively correlated if Pr(A∩B) > Pr(A)⋅Pr(B) and negatively correlated if Pr(A∩B) < Pr(A)⋅Pr(B).
In the case that $|\Omega|=p$ is prime and the space is equipped with the uniform probability density, one can check that each and every nontrivial event will be of the form $\frac{q}{p}$ with $\gcd(p,q)=1$ for some integer $q$ between $0$ and $p$ since the uniform probability distribution has the property that $Pr(A)=\frac{|A|}{|\Omega|}$ and prime numbers are by definition relatively prime to all integers positive integers less than it.
So then, supposing that $A$ and $B$ are in fact independent of one another, we have $A\cap B$ is an event with probability $\frac{|A|\cdot |B|}{p^2} = \frac{q}{p}$ for some $q$.
Multiplying both sides by $p^2$ we have $|A|\cdot |B|=p\cdot q$ and so $p$ divides evenly into $|A|\cdot |B|$ and as $p$ is prime divides into $|A|$ or divides into $|B|$, but since $A$ and $B$ are both nontrivial neither $|A|$ nor $|B|$ are equal to $0$ or $p$ but instead an integer strictly between $0$ and $p$... a contradiction.
Since Pr is the uniform probability distribution on $\Omega$, $Pr(A) = |A|/|\Omega|$, and $Pr(B) = |B|/|\Omega|$, so $Pr(A) \cdot Pr(B) = |A|\cdot|B|/|\Omega|\cdot|\Omega|$. Suppose A and B are independent for a contradiction, we have $Pr(A) \cdot Pr(B) = |A|\cdot|B|/|\Omega|\cdot|\Omega| = Pr(A \cap B) = q / |\Omega|\cdot|\Omega|$ for some positive q. Then we have $q \cdot |\Omega|= |A| \cdot |B|$. Since $|\Omega|$ is prime, $|\Omega|$ divides either $|A|$ or $|B|$ by Euclid's Lemma, but A and B are two non-trivial events, so $|A| \neq p \text{ or } 0$ and $|B| \neq p \text{ or } 0$. This is a contradiction.