How to calculate probablity for approaching destination for choosing path in directed graph

Question

[Scene:]

This is the problem...here A,B,C, and D represent path connected to each other Thsi represent the directed path shown by the arrow..

[problem:]

Here i have to find probablity that if someone start from 'A' then by what probablity it will approach 'D' via path 3 and 4 path...

[What I did..:]

There are following possible ways to approach D from A

1. A->C->D

2. A->B->C->D

3. A->B->D

Then i decided to find probablity for every possiblity..

(probablity of visiting C via all possible path from A to C)

There are total 3 possible path from 2,but only 2 satisfy for approaching C..

$$ \therefore p(A->C)=\frac 23$$

similarly..

$$p(A->B)=\frac 13$$

$$p(B->D)=\frac 12$$

$$p(C->D)=\frac 13$$

$$ Since we know,Selection of path on each destination is independent of selection made in previous destination.. Which means..They all are independent events...

Hence By applying product rule...

$$p(A->D)= p(A->C) \times p(A->B) \times p(B->D) \times p(C->D) $$ $$ =\frac 23 \times \frac 13 \times \frac 12 \times \frac 13 $$

$$=\frac 1{27}$$

This is the probablity of approaching A->D via path 3 or 4

[Question]

1.Have I approached right?

2.What will i get if I Sum up all probablity..What kind of event it wil reflect?

Answer 1

Let $e_1, e_2...e_n$ denote a sequence of edges, $e_i\in\{1,2...8\}$.

$p(1)=p(2)=p(8)=1/3$

$p(1,3)=p(1,4)=p(1,5)=p(1)\times\frac13=\frac19$

$p(2,3)=p(2,4)=p(2,5)=p(2)\times\frac13=\frac19$

$p(8,6)=p(8,7)=p(8)\times\frac12=\frac16$

$p(8,7,3)=p(8,7,4)=p(8,7,5)=p(8,7)\times\frac13=\frac1{18}$

It is easy to see that:

$p(1,3)+\ p(1,4)+\ p(1,5)+\ p(2,3)+\ p(2,4)+\ p(2,5)+\ p(8,6)+\ p(8,7,3)+\ p(8,7,3)+\ p(8,7,4)+\ p(8,7,5)=1=p(A\rightarrow D)$

$\mbox{$p(A\rightarrow D$ via $3/4)$}=p(1,3)+\ p(1,4)+p(2,3)+\ p(2,4)+\ p(8,7,3)+\ p(8,7,4)=\frac19+\frac19+\frac19+\frac19+\frac1{18}+\frac1{18}=\frac59$