tough probability problem

[davedave]

Consider the diagram below which consists of 3 horizontal lines and 4 vertical lines to form a grid. It shows a partial map of a certain city park with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C walking only to the south and east. Show that the probability that the tourist passes through point B is 3/5.


A-----
1 1 1 1
--- B--
1 1 1 1
------C

[tiny-tim]

… 3 horizontal lines and 4 vertical lines …

hihi davedave! :smile:

Your diagram shows 5 horizontal lines … what are the 1's? :confused:

Do you mean
A x x x
x x B x
x x x C ?

[scarecrow]

That's easy. Map out all the possible paths (from A-->B, south/east), and then count how many intersect B.

[davedave]

Consider the diagram below which consists of 3 horizontal lines and 4 vertical lines to form a grid. It shows a partial map of a certain city park with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C walking only to the south and east. Show that the probability that the tourist passes through point B is 3/5.


A-----
1 1 1 1
--- B--
1 1 1 1
------C

Hi tiny-tim,

Sorry for the confusion. The dashes represent horizontal lines and the "1" s are the vertical lines.

All my friends and I got 3/10 for the answer. We think there is a typo in the book's answer.

What do you think?

[Pere Callahan]

I think the answer is 3/5, which i got by mere counting.

How many paths do you think there are between A and C. How many of these do you think meet B?

[davedave]

I think the answer is 3/5, which i got by mere counting.

How many paths do you think there are between A and C. How many of these do you think meet B?

All my friends and I found 3 paths from A to B and 10 paths from A to C by counting. So,
our answer is 3/10. We don't see how to get 3/5.

[quadraphonics]

I count 10 distinct S/E paths from A to C, and 6 of these intersect B, so the answer is 6/10, or 3/5.

[davedave]

I count 10 distinct S/E paths from A to C, and 6 of these intersect B, so the answer is 6/10, or 3/5.

Consider the diagram of this problem which consists of 3 horizontal and 4 vertical lines.

A+++
++B+
+++C
In the problem, you can go ONLY south and east. If you go EAST from A to the
2nd "+" and down to B, there is only 1 path. If you go SOUTH from A to the 1st
"+" and move EAST to B, there are only 2 paths.
So, there are 3 paths from A to B and 10 paths from A to C.

Therefore, the probability from A to C is 3/10.

How can you get 6 paths from A to C?

[Pere Callahan]

All my friends and I found 3 paths from A to B and 10 paths from A to C by counting. So,
our answer is 3/10. We don't see how to get 3/5.

It is correct that there are 3 S/E paths from A to B, but that's not what you want to count. You want to count the number of S/E paths from A to C meeting B. So for each of your paths from A to B you have two possibilities to complete it to a A->C path. (Either going first East and then South, or first South, then East). This gives a total of 6 S/E paths from A to C meeting B.

I think your problme does not lie in counting paths itself but in determining what you want to count. In this case it's the numbers of ways to get from A to C via B. That means you have to account for all possibilities to get from A to B (which you did, obtaining three) but also for the possibilities to get from B to C (which is two and modifies your result to six.)

The total number of admissible paths joining A and C is right.