Deep Understanding of Independence of Probabilities I really want to have a deep understanding of the independent probabilities of two events.
That means to me that I just do not want to use and know the definition. I want to fully understand the why. 

Definition 3.1. (a) Two events $A$ and $B$ are independent if $\text{P}(A \cap B) = \text{P}(A)P(B)$.
  (b) $A$ (possibly infinite) collection of events $(A_i)_{i \in I}$ is an independent collection if for every finite subset $J$ of $I$, one has
  $$\text{P}\left(\bigcap_{i \in J} A_i\right) = \prod_{i \in J}\text{P}(A_i).$$
  The collection $(A_i)_{i\in I}$ is often said to be mutually independent.

Therefore my question:
Let's consider two events of two train crashes $A$ and $B$. $A$ is in London and $B$ is in New York.
If the intersection of the two trains is a multiple of the probabilities of the two events then these two events are independent.(In my opinion this should be $0$ for independent events) If not they are dependent.
If we just know this kind of information, logically these two events should not be dependent, because a train crash in London should not have anything to two with one in New York. (Could I also say shares the same information?) However, if I get that these two events are dependent is my equation wrong? AND is this value the probability of their dependency?
I appreciate your answer!
 A: There is a difference between mutually exclusive events and independent events (indeed, a very strict difference in the sense that a couple of non-trivial events cannot be both mutually exclusive and independent).
Events $A$ and $B$ are mutually exclusive if it is impossible for both of them to occur simultaneously. To borrow from your example, it is impossible for the same train to be involved in a crash in London (event $A$) and New York (event $B$) at the same time. So $\text{P}(A \cap B) = 0$. Another example is, if a die is rolled once, and $A$ and $B$ are the events of getting a $1$ and $6$ (respectively). Then either $A$ occurs, or $B$ occurs, or neither occurs. But both $A$ and $B$ cannot occur. So $\text{P}(A \cap B) = 0$.
Events $A$ and $B$ are independent if the occurrence (or non-occurrence) of one does not affect the (probability of) occurrence of the other. In other words, $B$ has the same probability of occurrence independent of whether $A$ has occurred or not, and vice versa. To borrow from your example again, the occurrence of a train crash occurring in London (event $A$) does not normally change the likelihood of a train crash (involving some other trains) in New York (event $B$). I say "normally", because of course, in the real world there are several factors at play which we may not know about. For example, if there is a lot of publicity about the train crash in London this week, it might very lower the likelihood of a train crash in New York in the coming few weeks (as people might take more precautions than usual).
So let's take a simpler example. If two dice are rolled at the same time, and $A$ and $B$ are, respectively, the events of getting a $1$ on the first die and a $6$ on the second, then $A$ and $B$ are independent (again, assuming there is no hidden connection between the dice via the atmospheric disturbances they create, gravitational waves, and whatnot!). Can both occur simultaneously? Certainly, why not? It is perfectly conceivable that the first die shows $1$ and the second shows $6$ (as I'm sure you've seen many times, roughly one-thirty-sixth of the time, in fact, if you've played Monopoly). So $\text{P}(A \cap B)$ is not $0$. For fair dice, $\text{P}(A) = 1/6$, and $\text{P}(B) = 1/6$. Then what is the probability that both $A$ and $B$ occur? It is simply $\text{P}(A)\text{P}(B) = 1/36$, because the events are independent.
Now let's look at events that are not independent. You are in Pall Mall right now, and you need to get to Marylbone Station in exactly two turns. What is the probability that this (let's call it event $A$) will happen? You need a $4$ to get to Marylbone Station, and that can happen in exactly two turns only if you get $2$ in the first as well as the second turn. So the probability of $A$ is the probability of getting the single combination $(2, 2)$ out of all the $36 \times 36 = 1296$ possible combinations, that is, $\text{P}(A) = 1/1296$. This is what you calculate before throwing the dice. Now you throw the dice and you get a $2$ (let's call this event $B$). Is the probability of $A$ still $1/1296$? Of course not. Now that you know that you've got a $2$ and reached Whitehall, you know that you only need a $2$ in the next turn to reach Marylbone Station, and this has probability $1/36$. Thus, given that $B$ occurred, the probability of occurrence of $A$ changes. We write it in this way $\text{P}(A|B) = 1/36$ (read "Probability of $A$ given $B$"). This is called conditional probability. Conditional probability of $A$ given $B$ is defined as $\text{P}(A|B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)}$. Thus, $\text{P}(A \cap B) = \text{P}(B)\text{P}(A|B)$ (this is usually called the Multiplication Theorem).
For independent events $A$ and $B$, $\text{P}(A|B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)} = \dfrac{\text{P}(A)\text{P}(B)}{\text{P}(B)} = \text{P}(A)$. This makes sense, as it essentially says that the probability of $A$ is the same whether $B$ occurs or not.
