Given a $p-$sided Dice, where $p$ is a prime number. Show that in a discrete model of a single throw of the dice $(\Omega, \mathcal{P}(\Omega), P)$:
Events $A, B$ are independent $\iff$ At least $A$ or $B$ is $\Omega$ or is $\emptyset$
I am able to prove the trivial "$\Leftarrow$" here:
Let $B=\Omega$
By definition: $P(A|B)=\frac{P(A \cap B)}{P(B)}$. Since $B=\Omega, A\subset B$ and thus $A \cap B = A$, while $P(B)=1$, and thus we get $P(A|B)=P(A)$.
It could go on in a similar fashion for $B= \emptyset$.
I get into problems on "$\Rightarrow$", as I do not know how to incorporate the fact that the dice is "Prime-Sided"
I initially wanted to go about it using contradiction, assuming $A, B \neq \emptyset, \neq \Omega$. But I cannot find any contradictions, I assume it has something to do with the prime numbers.
An event $A$ in this model is a subset of ${\mathbb Z}_p$ containing $n_A\in[0\,..\,p]$ elements. Its probability is ${n_A\over p}$. Let two events $A$ and $B$ be given, and assume $\#(A\cap B)=m$. If $A$ and $B$ are independent then we have $${m\over p}={n_A\over p}\cdot{n_B\over p}\ ,$$hence $n_A\cdot n_B= p\cdot m$. Now draw conclusions.