On thinking about independent events The definition of indepedent event is the following,
$P(A \cap B) = P(A)P(B)$.
Via the conditional dependence definition this makes some kind of sense. Is there another way to think about this equality? I been staring at it for a while but I cant seem to get any intuition except via the conditional.
Any ideas?
 A: Suppose that two events $A$ and $B$ are independent. Let's assume, for instance, that $A$ takes place one in each four occasions, whereas $B$ takes place one in each nine occasions. Then how often do they both occur? In each $36(=4\times9)$ occasions, $A$ occurs about $9$ times and $B$ occurs about $4$ times. If these times are distributed more or less uniformly, it is to be expectes that $A$ and $B$ occur simultaneously only once. That is, $A$ and $B$ occur simultaneously $1$ in each $36$ occasions.
A: The clearest way of understanding the notion of independence is certainly in terms of the conditional:
$$
P(A\mid B) = P(A).
$$
That is, knowledge of $B$ doesn't affect the probability of $A$.
But all hope is not lost upon using the definition $P(A\cap B)=P(A)P(B)$. This means that when trying to find the probability that both $A$ and $B$ occur, we need only find the probabilities of $A$ and $B$ "independently" and take their product. For instance, say you flip five coins and you want to find the probability that they are all heads. It's intuitive that, in order to do this, we find the probability that each is heads and take their product. This is because the probability that the second coin is heads (presumably) doesn't depend on whether the first was heads, and the third doesn't depend on the first two, and so forth.
A: If you think of a probability as the fraction of the time something happens, it is conceptually clear that the probability that two things happen must be no greater than the probability that either one of them happens individually.  So $\Pr[A\cap B]\le \Pr[A]$ and $\Pr[A\cap B]\le\Pr[B]$.
The question is, in what way does $\Pr[A\cap B]$ get reduced relative to $\Pr[A]$ or $\Pr[B]$?  Let's say $\Pr[A]=\frac{1}{3}$ and $\Pr[B]=\frac{2}{5}$, so that $A$ happens $\frac{1}{3}$ of the time and $B$ happens $\frac{2}{5}$ of the time.  If one didn't think about it carefully, one might think that $B$ happens on $\frac{2}{5}$ of those occasions when $A$ happens and therefore that $A\cap B$ happens $\frac{1}{3}\cdot\frac{2}{5}=\frac{2}{15}$ of the time, in just the same way as $\frac{2}{5}$ of $\frac{1}{3}$ of a cake is $\frac{2}{15}$ of a cake.   But that is, in fact, only true when $A$ and $B$ are independent.  It could well be that $B$ happens whenever $A$ happens, so that $\Pr[A\cap B]=\Pr[A]\cdot1=\Pr[A]$.  Or it could be that $B$ never happens when $A$ happens so that $\Pr[A\cap B]=\Pr[A]\cdot0=0$.  In fact, $\Pr[A\cap B]$ in this example can take any value between $0$ and $\Pr[A]$ depending on whether the occurrence of $A$ makes $B$ less likely, leaves the probability of $B$ unchanged, or makes $B$ more likely.
Of course this discussion really hasn't got away from the conditional definition.  The general formula for $\Pr[A\cap B]$ is
$$
\Pr[A\cap B]=\Pr[A]\cdot\Pr[B\mid A].
$$
Independence is precisely the condition that the occurrence of $A$ makes $B$ neither more nor less likely, $\Pr[B\mid A]=\Pr[B]$, so that the formula becomes $\Pr[A\cap B]=\Pr[A]\cdot\Pr[B]$.  Nevertheless, I hope that this way of thinking about it makes the multiplicative definition of independence just as intuitive as the conditional definition.
Added note: I have changed the numbers in the example above because the lower bound on $\Pr[A\cap B]$ was false using the original numbers.  In general, the truth is slightly more complicated: if $\Pr[B]\ge\Pr[A]$ then $\Pr[A\cap B]$ can take any value between $\max(0,\Pr[A]+\Pr[B]-1)$ and $\Pr[A]$.  My original numbers were $\Pr[A]=\frac{3}{5}$ and $\Pr[B]=\frac{2}{3}$.  Since the sum of these probabilities is greater than $1$, it is not possible for $A$ and $B$ to be disjoint, i.e. for $\Pr[A\cap B]$ to be $0$.  In this example the intersection is as small as possible when $\Pr[A\cap B]=\Pr[A]+\Pr[B]-1$, which is where the lower bound in this case comes from.  So for these numbers, $\Pr[A\cap B]$ must lie between $\frac{3}{5}+\frac{2}{3}-1=\frac{4}{15}$ and $\Pr[A]=\frac{3}{5}$.
A: Since $P\left(A/B\right)=\frac{P(A\cap B)}{P(B)}$,  you can write it in the following form.
$$P\left(A/B\right)=P(A).$$
Now, I think it's clear.
I think the following example will help. 
We know that the birth of children is independent from the previous birth.
Assume that there are two births and 
$A$ is an event that was born girl in the first time, while
$B$ is an event that was born boy in the second time.
$$\Omega=\{BG,BB,GB,GG\},$$
$$A=\{GG,GB\},$$
$$B=\{GB,BB\}$$ and
$$A\cap B=\{GB\}.$$
Thus, $$P(A)=\frac{2}{4}=\frac{1}{2},$$
$$P(B)=\frac{2}{4}=\frac{1}{2},$$
$$P(A\cap B)=\frac{1}{4}$$ 
and we see that indeed
$$P(A\cap B)=P(A)P(B).$$
A: Too many formulas for what I consider intuition.
The big question is, where does the multiplication come from?
Let's say that we have a trustworthy weather report that says the coming week every day has a 50% chance of rain.
Let's flip a coin each day.
What is the chance of the coin coming up heads on a days it rains?
To observe all possible outcomes, it is easy to see I need to distinguish 4 possibilities: it rains or not, and the coin comes up heads or not.
With that 50/50 chance for each event, all four possibilities are intuitively equal, so the chance of rain ∩ heads = 1/4. But of course, I do not immediately have to see that that is equal to 1/2 * 1/2.
Now what happens if the weather report says the chance for rain is only 25%?
P(heads) is still 1/2, but my four options are obviously not equal anymore.
Actually, if I look at any 100 days I toss heads, I have expect only 25 of them to be rainy. So out of the coming 200 days I toss a coin, only 25 are expected to be rainy days with a heads toss. 
Some pattern must start emerging: Of any day I toss a coin, P(heads) result in heads, and of those P(rain) are rainy heads days. That, for me at least, quite quickly leads to seeing that P(heads) and P(rain) can simply be multiplied to get to P(rainy day and tossing heads).
