Conditional Probability in Pebble World Definition of the Pebble World (taken from Stat 110): In the Pebble World, the definition says that probability behaves like mass: the mass of an empty pile of pebbles is $0$, the total mass of all the pebbles is $1$, and if we have non-overlapping piles of pebbles, we can get their combined mass by adding the masses of the individual piles. The pebbles can be of differing masses and we can also have a countably infinite number of pebbles as long as their total mass is $1$.


In the above image, suppose each pebble weighs $\frac{1}{9}$ (for brevity). The pebbles inside the red box denote event $A$ and the pebbles inside the green box denote the event $B$. Note that the two boxes are intersecting.
What is $P(A | B)$? It's supposed to be the probability of $A$ given $B$ has occurred, which means, $\frac{2/9}{6/9} = \frac{1}{3}$. How do we explain this intuitively? ($1/3$ feels like selecting the one element that's not in the intersection upon all the elements inside the red box. I don't think it's meant to be this way.)

Let:

*

*the sample space (picking any pebble) be the black square;

*$A$ be event of picking any pebble within the red rectangle;

*$B$ be event of picking any pebble within the green rectangle.

Say, we know that event $B$ occurs; this means that $A$ can possibly eventuate from within the green rectangle. In other words, being given that $B$ occurs is the same as narrowing down the effective sample space from the black square to the green rectangle.
In this case (i.e., given that $B$ occurs) the probability of $A$ $$=\frac{\text n(\text{‘success’})}{\text n(\text{the effective sample space})}\\=\frac{\text n(\text{the part of the red rectangle that's inside the green rectangle})}{\text n(\text{the green rectangle})}\\=\frac{\text n(A\cap B)}{\text n(B)}.$$
In other words, $$P(A|B)=\frac{\text n(A\cap B)}{\text n(B)}\\=\frac26\\=\frac13.$$
