What are independent events? I know that if events $A$ and $B$ are independent, then $P(A\cap B) = P(A).P(B)$.
Lets say I roll a die, and define the following events:
$x$ = Number is a multiple of $3 = \lbrace 3, 6\rbrace$
$y$ = Number is more than $3 = \lbrace 4, 5, 6\rbrace$
$z$ = Number is even  =  $\lbrace 2, 4, 6 \rbrace$
Therefore,
$P(x) * P(y)$: 0.166666666667
$P(x \cap y)$   : 0.166666666667
$P(x) * P(z)$ : 0.166666666667
$P(x \cap z)$    : 0.166666666667
$P(y) * P(z)$ : 0.25
$P(y \cap z)$    : 0.333333333333
My question is, why are $xy$ and $xz$ independent events but $yz$ is not? I had expected all combinations of events to be independent as neither events affect any other.
 A: To add to the Rob's answer the joint probability of two events in the discrete case can be decomposed as follows (See the wiki's definition for the discrete case) :
$P(A \cap B) = P(A|B) P(B)$
Therefore, if the fact that we observe the event $B$ is not informative about the probability of our observing the event $A$ then it must be the case that:
$P(A|B) = P(A)$.
Thus, if the two variables are independent (i.e., the occurrence of one event does not tell us anything about the other event) then it must be the case that:
$P(A \cap B) = P(A) P(B)$
The crucial condition for independence of two events is: $P(A|B) = P(A)$. This condition is essentially stating that the "Probability of our observing event A over the unrestricted sample space = Probability of our observing the event A if the sample space is restricted to the set of possible outcomes compatible with the event B". Thus, the knowledge that the event B occurred has no information about the probability of observing the event A.
A: I find it easiest to think of independence in terms of conditional probabilities: $X$ and $Y$ are independent if $P(Y|X)=P(Y)$ and $P(X|Y)=P(X)$. That is knowing one does not change the probability of the other. In this example,
\begin{aligned}
P(Y|X) &= P(Y \cap X) / P(X) = 1/2 = P(Y)\\
P(Y|Z) &= P(Y \cap Z) / P(Z) = 2/3 \ne P(Y)\\
P(Z|X) &= P(Z \cap X) / P(X) = 1/2 = P(Z)
\end{aligned}
So knowing $X$ does not affect the probability of either $Y$ or $Z$. But knowing $Z$ does affect the probability of $Y$.
