Independence of two events with non empty intersection Let $A, B$ be two non empty events. If they are disjoint, i. e. exclusive, they are not independent. In the case they are not  disjoint, they can be either independent or not independent. Intuitively, in the case two events have non empty intersection, the occurrence of one event will condition the occurrence of the other one since their intersection is not empty. In other words, I don't understand how $P(A|B) = P(A). $I am thus confused. 
Can you provide an explanation and an example of 2 non empty events with non empty intersection which are independent? 
Many thanks. 
 A: Take a 4-sided die with the numbers 1 to 4. Check the events:


*

*result is $\le 2$.

*result is odd.
Another possibility: if $P(A) = 1$ and $B\subset A$ then
$$P(A) P(B) = P(B) = P(A\cap B).$$
A: Independence $P(A|B) = P(A)$ is the math way of saying that knowing $B$ does not affect the probability of $A$.  E.g.


*

*$A =$ whether a first fair coin flip is Heads

*$B =$ whether a second fair coin flip is Heads (assuming these are "normal" coins that don't affect each other)
So $P(A) = P(B) = P(A|B) = P(B|A) = 1/2$ while $P(A\cap B) = 1/4$.
Perhaps this is where you're confused: Conditioned on $B$ does "shrink" the sample space (to the subset which is $B$), so it does potentially "shrink" the ways how $A$ can happen (or not happen).  However, independence means the "shrinkages" are "exactly proportional" in some sense, so that it doesn't change the probability of $A$ happening.  In this specific example: 


*

*The full sample space is $\{HH, HT, TH, TT\}$ (where the first symbol denotes $A$'s result and the second symbol $B$'s result)

*Conditioned on $B$ "shrinks" the space to $\{HH, TH\}$

*However, the prob of $A$ happening was $\frac24 = \frac12$ "before" and is still $\frac12$ "afterwards"
A: Consider the Karnaugh table for 2 events $A, B$ and their complementary events $A^C,B^C$ with, in the boxes, the different probabilities of possible events :

Saying that $A$ and $B$ are independent is thus expressed in the following way (under the constraint $x+y+z \leq 1$) :
Rational or real values of $x,y,z$ are such that :
$$P(A \cap B)=P(A) \times P(B) \ \ \ \iff \ \ \ x=(x+y)(x+z)$$
which can be realized in an infinite number of ways...
