What does multiplication mean in probability theory?
I like this answer taken from http://mathforum.org/library/drmath/view/74065.html :
" It may be clearer to you if you think of probability as the fraction of the time that something will happen. If event A happens 1/2 of the time, and event B happens 1/3 of the time, and events A and B are independent, then event B will happen 1/3 of the times that event A happens, right? And to find 1/3 of 1/2, we multiply. The probability that events A and B both happen is 1/6.
Note also that adding two probabilities will give a larger number than either of them; but the probability that two events BOTH happen can't be greater than either of the individual events. So it would make no sense to add probabilities in this situation. "
If you randomly pick one from $n$ objects, each object has the probability $\frac{1}{n}$ of being picked. Now imagine you pick randomly twice - one object from a set of $n$ objects, and a second object from a different set of $m$ objects. There are $n\cdot m$ possible pairts of objects, and thus the probability of each individual pair is $\frac{1}{n\cdot m} = \frac{1}{n}\cdot \frac{1}{m}$.
More generally, let $A$ be some event with probability $\Pr(A) = a$, and $B$ some other event with probability $\Pr(B) = b$. Assume you already know that $A$ happened, meaning that instead of looking at the whole probability space (i.e. at the whole set of possible outcomes), we're now looking at only $A$. What can we say about the probability that $B$ happens also, i.e. about the probability $\Pr(B\mid A)$ (to be read as "the probability of $B$ under the condition $A$")?
In general, not much! But, if $A$ and $B$ are independent, then by the definition of independence, knowing that $A$ has happened doesn't provide us with any information about $B$. In other words, knowing that $A$ has happened doesn't make the likelyhood of $B$ happening also any smaller or larger, so $$ \Pr(B\mid A) = \Pr(B) \text{ if $A,B$ are independent.} $$
Now look at $\Pr(A \cap B)$, i.e. the probability that both $A$ and $B$ happen. We know that if $A$ has happened, then $A \cap B$ happens with probability $\Pr(B\mid A)$. If we don't know that $A$ has happened, we have to scale this probability with the probability of $A$. Thus, $$ \Pr(A \cap B)= \Pr(B\mid A)\Pr(A) \text{.} $$ [ You can imagine $A$ and $B$ to be some shapes, both inside some larger shape $\Omega$. $\Pr(A\cap B)$ is then the percentage of the area of $\Omega$ that is covered by both $A$ and $B$, $\Pr(A)$ the percentage of the area of $\Omega$ covered by $A$, and $\Pr(B\mid A)$ is the percentage of the area of $A$ covered by $B$. ]
If $A,B$ are independent, we can combine these two results to get $$ \Pr(A\cap B) = \Pr(A)\Pr(B) \text{.} $$
You are still scaling, but by numbers that are smaller than $1$. In your example, you are scaling $1/2$ by a factor of $1/2$, scaling it down to $1/4$. The first $1/2$ represents the outcomes where the first coin flipped is heads. But only $1/2$ (the second "$1/2$" from your example) of those outcomes also have the second coin come up heads.