I'm trying to understand independence between events. I have a couple examples that I could use guidance on. Specifically, if my reasoning is correct or if there is something else I should consider.
I use first the two independent events of rolling a 6 with a fair die.
$DIE = \{1,2,3,4,5,6\}$, $S=DIE \times DIE$
$|S| = 6 \times 6$
$A_0 = \{(x, y) : x = 6, y \in DIE\}$, $A_1 = \{(x, y) : x \in DIE, y = 6 \}$, $A_0 \cap A_1 = (6, 6)$
$P(A_0) = |A_0|/|S| = (1 \times 6)/(6 \times 6) = 1/6$, $P(A_1) = |A_1|/|S| = (6 \times 1)/(6 \times 6) = 1/6$, $P(A_0 \cap A_1) = 1/36$
So we check for independence: $P(A_0 \cap A_1) = P(A_0)P(A_1) = 1/36 = 1/6*1/6 = 1/36$
So $A_0$ and $A_1$ are independent. My first question is there a better way of spotting the independence? For instance does specifying the ordered pair to be the entire sample space for that order cause this property? Wiki says independence is "if the occurrence of one (event) does not affect the probability ... the other".
The second half of my question is an example of two non-independent events. The first event is getting a 6 on the first roll of the die and the event that the first and second rolls sum to 8.
I'm guessing the criteria for non-dependence is because the second event depends on the 6 in the first event, but this confuses me because similar occurs in the above example.
I reuse $A_0$ from above and specify $A_2$.
$A_2 = \{(x,y): x + y = 8\} = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$
$|A_2| = 5$
$A_0 \cap A_2 = \{(6, 2)\}$
$P(A_0 \cap A_2) = 1/36$
$P(A_0)P(A_2) = 1/6 * 5/36$
$P(A_0 \cap A_2)\ne P(A_0)P(A_2)$, therefore non-independent but why?