Multiplication Rule of Independent Events I am reaching to this community because I know that the answers to the apparently simple question I have would be as substantial and involved as the core mathematical analysis of statistical formulas. I am in search of an intuitive explanation of why the multiplication rule of independent events works.
If A and B are independent events, 
P(A∩B)=P(A)×P(B)
What is the reasoning behind this rule? 
 A: Let's start from conditional probability, which is fairly intuitive.
Suppose we are tossing a fair coin twice. Because a coin with memory would
be hard to explain, defining $A = \{\text{Heads on 1st toss}\}$ and 
$B = \{\text{Heads on 2nd toss}\}$, It seems that the probability of $B$ should not depend on the occurrence of $A$.
Perhaps
write this as $P(B|A) = P(B).$ The probability that $B$ occurs is the
same, whether or not we know that $A$ occurred. "Event $B$ is independent of event $A.$" 
Somewhat expanded, this is
$$P(B|A) = \frac{P(A \cap B)}{P(A)} = P(B).$$
Which leads to
$$P(A \cap B) = P(A)P(B),$$
and even to $P(A|B) = P(A),$ which amounts to "$A$ independent of $B.$" 
[If you don't see the result of the first toss, does knowing the second was Heads influence
your bet whether the first toss resulted in Heads?]
So we say the events are "independent of each other," a symmetrical
statement. Of the equations above the only symmetrical one is 
$P(A \cap B) = P(A)P(B).$ So it is convenient to use that as the 'definition'
of independence. [Incidentally, this definition also gets rid of any
concern about having $0$ probabilities in denominators.]
This definition works 'two' ways: (a) If you feel intuitively that $A$ and $B$
are independent, then your probability assignment has to be built consistent
with the equation $P(A \cap B) = P(A)P(B).$ In our case, this can be done
by assigning probability $1/4$ to each of the four outcomes in the
sample space, which can be written $S = \{\text{HH, HT, TH, TT}\}.$ 
Checking: That assignment implies $P(A) = P\{\text{HH. HT}\} = 1/4 + 1/4 = 1/2.$
Similarly, $P(B) =  P\{\text{HH. TH}\} = 1/2$ and $P(A \cap B) = P\{\text{HH}\}
= 1/4.$ And finally, by multiplication, $P(A \cap B) = P(A)P(B).$ 
(b) If someone gives you a probability model (assignment of probabilities
in the sample space), and that model implies $P(A \cap B) = P(A)P(B),$
then $A$ and $B$ are decreed to be independent events--whether or not that seems intuitive to you. One example, with an unfair coin, might be to assign probabilities $(1/9,\, 2/9,\, 2/9,\, 4/9)$ to
the respective elements of  $S = \{\text{HH, HT, TH, TT}\}.$ 
