Independence of an event and the sample space I have a problem for a homework that says A and B are events. It then requires me to decide if A and Ω (the sample space) are independent. 
My initial guess is to say yes, because if the sample space occurs, that should not influence the probability that A will occur on the next trial. 
My proof for this is as follows:
P(Ω & A) = P(A) * P(Ω|A) where P(Ω|A) = P(A)
so P(Ω & A) = P(A) * P(A)
but if they were independent then P(Ω & A) = P(A) * P(Ω) 
and since P(Ω) = 1, 
P(Ω & A) = P(A) and not P(A) * P(A) meaning they are not independent. 
So is my proof correct, or my intuition correct? Very confused right now and would appreciate some guidance. 
 A: Your intuition is not quite right. To say that $A$ and $B$ are independent events does not say anything about the probability that $B$ will occur on the next trial. Independence of events involves only one trial. I make one observation. Given that $A$ occurred on that trial, what is the probability that $B$ occurred on the same trial?
When $B=\Omega$, knowing that $\Omega$ occurred does not change the probability that $A$ happened since knowing $\Omega$ occurred gives you no information at all (it always occurs). Intuitively, this is why $\Omega$ and $A$ are independent.
Your proof is not correct. $P(\Omega\mid A)$ is not equal to $P(A)$ in general. It's also worth noting that it's not quite correct to just say "we get $P(A)^2$ instead of $P(A)$ therefore this is wrong", since it could be that $P(A)^2=P(A)$, for your reasoning to be complete you would need to mention that $x^2=x$ is only true when $x$ is $1$ or $0$.
A correct proof is to write down the definition of independence for $\Omega$ and $A$:
$$P(\Omega\cap A)=P(\Omega)P(A)$$
and note that $\Omega\cap A=A$ as sets and $P(\Omega)=1$, so that the equation reduces to $P(A)=P(A)$.
