I have a question about a pairwise independence example:
A die is rolled twice. Let $A$ be the events that the first roll is a $1$, $B$ be the event that the second roll is a $1$, and $C$ be the event that the sum is even.
The three events are not mutually independent, but they are pairwise independent.
If $C$ is the event that the sum is even, then clearly in depends on either, and both, $A$ and $B$, right? So how is it pairwise independent with $A$ and $B$? Thank you.
Write a pair of numbers $(x,y)$, with $x$ representing the number you get on the first roll while $y$ is the number you get on the second roll. You should get a total of 36 possibilities.
You can easily see that $P(C) = \frac{18}{36} = \frac{1}{2}$.
$P(A) = \frac{1}{6} = P(B)$. Clearly, the number you get on the first roll does not affect what you get on the second roll.
We have: $ P (A \cap B) = \frac{1}{36} = \frac{1}{6} \cdot \frac{1}{6} = P(A)P(B)$.
$P (A \cap C) = P(B \cap C) = \frac{3}{36} = \frac{1}{6} \cdot \frac{18}{36} = P(A)P(C) = P(B)P(C)$.
But $P(A \cap B \cap C) = \frac{1}{36} \neq \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{18}{36} = P(A)P(B)P(C)$. If this last part was also an equality we would have the mutually independent. So, in this case we only have the event being pairwise independent, as it was promised.