How to understand independence of probability? By definition, when $$P(E\,|\,F) = P(E)$$ holds, we say that $E$ is independent of $F$. 
By definition of conditional probability, $$P(E\,|\,F) = {P(E \cap F) \over P(F)} \Rightarrow P(E \cap F) = P(E)P(F).$$
I'm confused here that shouldn't $P(E \cap F) = 0$ if they are independent? 
 A: The two concepts of "independent" and "mutually exclusive" are different.
Two events are "mutually exclusive" ( that is, $$P(E \cap F) = 0$$) if they can't both happen at the same time. For example, if I roll a die and define E = "I roll a 6" and F = "I roll a 3", these can't both be true. If E happens, i know F didn't.
Two events are "independent" (that is, $$P(E \cap F) = P(E)P(F)$$) if the outcome of each has no influence at all on the other. For example if we each roll a die and define E =  "I roll a 6" and F = "you roll a 3". Whether E happens or not, it makes no difference to the probability of F happening.
The only way a pair of events can be both "mutually exclusive" and "independent" is if at least one of them has zero probability.
(Note: as noted by wythagorus below, this is not the same as completely impossible - eg any exact result in a continuous distribution is possible but with zero prob)
A: Intuitively, saying that two logical propositions $E$ and $F$ are independent means that learning whether $E$ is true or false tells you nothing about $F$, and vice versa.
For example, suppose you flip two coins, a penny and a nickel, and consider the propositions


*

*$E$ = "The penny came up heads"

*$F$ = "The nickel came up heads"


For ordinary coins, it's very reasonable to assume that $E$ and $F$ are independent, because learning whether or not the penny came up heads should tell you essentially nothing about whether the nickel did. If you come up with a probability model where $E$ and $F$ are not independent, your model is saying something very strange about how these coins behave.

There are four possible outcomes for the coin flips, which can be listed in a truth table, using $1$ for "true" and $0$ for "false":
$$\begin{array}{r|rrr}
E & 1 & 1 & 0 & 0 \\
F & 1 & 0 & 1 & 0
\end{array}$$
A typical statistical model for a pair of coin flips is to assume that each of the four possible outcomes is equally likely. In other words,
$$\begin{align*}
P(E \cap F) & = \tfrac{1}{4} &
P(E \cap \neg F) & = \tfrac{1}{4} &
P(\neg E \cap F) & = \tfrac{1}{4} &
P(\neg E \cap \neg F) & = \tfrac{1}{4}.
\end{align*}$$
In this model, are $E$ and $F$ independent?
It's not hard to calculate from the probabilities above that $P(E)$ and $P(F)$ are both $\tfrac{1}{2}$, and $P(E \mid F)$ is also $\tfrac{1}{2}$. Hence, $E$ and $F$ are independent. This is good, because we argued earlier that $E$ and $F$ should be independent in any reasonable model for coin flips.
Since $E$ and $F$ are independent, the calculation in your question tells us that $P(E) P(F)$ should be equal to $P(E \cap F)$. Indeed, $\tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$. Your condition $P(E) P(F) = P(E \cap F)$ is actually equivalent to independence: if the probabilities of two logical propositions satisfy your condition, the propositions are independent.
A: You're correct that if $E$ and $F$ are independent, then $P(E \cap F)=P(E)P(F)$. If $E$ and $F$  are independent, then it means that the result of $E$ has no influence on the result of $F$ and vice versa. For example:
I toss a quarter and a dime. The result of the quarter toss has no influence on the dime toss.
However, if $P(E \cap F)=0$, then they are mutually exclusive.
This is for example: I toss one quarter.


*

*$E : $ It is a head.

*$F : $ it is a tail.


They cannot happen at the same time. 
