A tough path-counting problem?

Question

Consider the diagram below. It consists of three horizontal and four vertical lines. Suppose that you start at point A and randomly select a path to point C walking only to the south and east. What is the probability that you pass through point B?

A + + +

+ + B +

+ + + C

Answer 1

6 / 9 im sorry

there are 9 possible  routes if u go only south and east

1. South, South, East, East, East (u get to C but do not passB)

2. South, East, East, South, East. (u passed B)

3. South, East, East, East, South (passed B)

4. East, South, South, East, East (did not pass B)

5. East, South, East, South, East (passed B)

6. East. South, East, East, South (passed B)

7. East, East, South, South, East (Passed B)

8. East, East, south, east, south (passed B)

9. East, east, east, south, south (did not pass B)

so out of all the 9 routes only 6 passes B

6/9 or one third

Answer 2

Count the number of paths from A to B, and the number of paths from B to C.  Then the probability of a path from A to C going through B is

(paths from A to B) * (paths from B to C) / (paths from A to C)

The number of paths between two points separated by X east-west units and Y north-south units is the same as the number of different ways you can flip X+Y coins and get X heads.

(Philip's answer of 6/8 is incorrect.)

(Philip's answer of 6/9 is also incorrect.)