How does the multiplication theorem correspond to the concept of intersection? Given two events A and B defined on a sample space S.
S : Rolling a six-sided dice
A : Getting an even number
B : Getting a number ≥ 4
In an elementary sense (the experiment being carried out once), intersection refers to an outcome which satisfies both the events.
For our example,
A ∩ B = {4,6}
Hence P(A ∩ B) = 1/3
However, when we think about the multiplication theorem, we see that the experiment is carried on twice.
(We roll two six-sided dice, one after the other, and then calculate the probability of obtaining an even number on the first and a number ≥ 4 on the second)
P(A ∩ B) = P(A) × P(B)
P(A ∩ B) = 1/4
My problem lies in the fact that both of them are represented as A ∩ B.
(Which I think is a little bizarre give that in the first case, we carry out the experiment once, and in the second case, we carry it out twice)
To summarize my query - 
How does intersection, in the elementary sense, in which we perform the experiment once, correspond to the intersection we obtain from the multiplication theorem, in which we carry out the experiment twice? 
 A: The experiment, of rolling two dice one after the other, is carried out once. The sample space is the set of ordered pairs
$$S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),\dots,(6,4),(6,5),(6,6)\}.$$
$$A=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}.$$
$$B=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)\}.$$
$$A\cap B=\{(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)\}.$$
A: The answer is that the two are not in general the same. You cited the multiplication theorem wrongly.$\def\pp{\mathbb{P}}$ It only applies for independent events $A,B$ that $\pp(A \cap B) = \pp(A) \pp(B)$.
In your example, $A,B$ are indeed not independent, and your calculations already show you that if $\pp(A \cap B) = \pp(A) \pp(B)$ then we would get $\frac{1}{3} = \frac{1}{4}$ which is a contradiction.
What happens when $A,B$ are independent is that in fact we can look at $A \cap B$ as either of the following:


*

*the outcome being in both of $A,B$.

*the same random process done twice, first outcome in $A$ and second outcome in $B$.
Since $A$ and $B$ are independent, in (1) knowing that the outcome is in $A$ gives no bias whatsoever to whether the outcome is in $B$ or not, and so it is no different from discarding the outcome, doing the random process again and testing whether the second outcome is in $B$. This is why (1) has the same probability as (2).
