Roll $2$ dice.
Let $E$ be the event that the sum of the dice is $6$
Let $F$ be the event that the sum of the dice is $7$
Let $G$ be the event that the first die rolled is a $4$
$E$ and $G$ are dependent (since $P(E\cap G) \neq P(E)P(G)$ )
$F$ and $G$ are independent (since $P(F\cap G) = P(F)P(G)$ )
Intuitively, why is this true?
Best Answer
Roughly speaking, if you know the sum of the dice is $6$, that gives you some information about what the first die rolled: namely, that it couldn't possibly have rolled a $6$. On the other hand, if you know the sum of the dice is $7$, that doesn't tell you anything about the result of the first die. For any number $n$ it might have rolled, the second die could have rolled the number $7-n$. So events relating to the result of the first die are independent of whether you rolled a $7$, but not of whether you rolled a $6$.
You can make this a little more rigorous by thinking in terms of conditional probabilities: if $A$ and $B$ are events with nonzero probability, then they're independent if and only if $P(A|B)=P(A)$ (or $P(B|A)=P(B)$). For your events, a bit of case analysis will show you that $P(G|F)=1/6=P(G)$, but $P(G|E)=1/5 \neq P(G)$.