Probability Trees Why do you multiply the probability of one branch by another to get the probability of a multiple-branch result. For example: I have three main branches, each of those has three branches, and each of those has two branches. My probability of ending up on any one is $\frac{1}{18}, or \frac{1}{3} \times \frac{1}{3} \times \frac{1}{2}$.  Or, I can just count all of the nodes at the end of the branches. 
 A: It sounds like you're asking why you multiply through events when examining situations in conditional probability. 
Conditional probability is defined as
$$ P(A\vert B) = \frac{P(A\cap B)}{P(B)} $$
And, for those who aren't familiar with the notation, reads off as
"The probability of Event A given that Event B has occurred equals the probability of the intersection of Events A and B divided by the probability of Event B." 
If we multiply both sides by $ P(B) $, we get 
$$ P(A\cap B) = P(A\vert B) P(B) $$
This can be represented graphically as:

Which would solve our problem for a tree diagram for a two branch diagram. However, what would we then do if we wanted to go further than this? What if I had a finite amount of sets $N$ and I wanted to find the probability of some event $A$ given that all sets $B_1, B_2, ... , B_N$ had occurred? 
First, let us assume that all of these events $B_1, B_2, ... B_N$ are mutually exclusive, that is, $B_1 \cap B_2 \cap .... B_N = 0$. This means that at any given time, we can only have one event occur. An example of this would be $B_1$ = turning right or $B_2$ = turning left. You can't turn left and turn right at the same time; only one can happen at a given time. 
We also need to assume that all of these sets $B_1, B_2, ... , B_N$ are exhaustive; that is, $B_1 \cup B_2 \cup .... B_N = 1$. This means that for a given event $A$, there are $N$ "paths" of event $B$ that could have caused this $A$. Another way of thinking of it is to go back to the turning left/turning right example. Imagine you are on a fork on the road; there are two paths, one goes right, another goes left. To travel forward, you must take one of the two paths. Thus, in our example, if we denote travelling forward as $A$, then there are $N=2$ "paths" of achieving $A$, $B_1$ being taking the right fork, $B_2$ being taking the left fork, and the union of all events $B$, in our case, $B_1 \cup B_2 = 1$, thus describing every possible way of travelling forward.
A graphical way of representing this would be as follows: 
 
Where each color is a different event $B_i$, and the total circle formed is $A$.
We can then say that 
$$ P(A) = P(A\cap B_1)P(B_1)+P(A\cap B_2)P(B_2)+...P(A\cap B_N)P(B_N) $$ 
$$ P(A) = \sum\limits_{i=1}^N P(A\cap B_i)P(B_i)$$
Which is known as the Law of Total Probability.
Now, you're probably wondering, how the heck does this connect to my original question? After all, we're looking for some weird event now and you're slamming me with all of this set notation and some weird "Total Probability" stuff, man, where's the application to the original question? Let me show you!
Here's a picture I made of what's going on in the question asked:

So, let's say we're looking to get to node P in our diagram. 
Now, to get to P, we must have also gotten to B and F. Thus, we can express the probability of reaching P as 
$$ P(P) = P(P\cap F\cap B) = P(P \vert F\cap B)P(F\cap B) $$
And by the definition of conditional probability, where we showed earlier that 
$$ P(A\cap B) = P(A\vert B) P(B) $$
was true, we can then expand our expression to 
$$P(P) = P(P \vert F\cap B)P(F\vert B)P(B)$$
Where
$$P(B) = \frac{1}{3} $$
$$P(F\vert B) = \frac{1}{3} $$
$$P(P \vert F\cap B) = \frac{1}{2} $$
So
$$P(P) = \frac{1}{18}$$
Alternatively, we could have used the Law of Total Probability to calculate our probability.
By the Law of Total Probability
$$ P(P) = \sum\limits_{i=1}^N P(A\cap B_i)P(B_i) $$
Which equals 
$$ \frac{18}{18} $$
As it is the summation, where N=18, of the probability of going to any node, which we just calculated.
And as we are looking to get to node P, we just divide that by 18 (as there are 18 total nodes) to get $\frac{1}{18}$ as before.
Hope this helped!
Edit: It looks like I forgot a third branch for the B node - my bad! Just pretend it's like the others, hahahaha.
