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Tough probability problem

  1. Aug 25, 2008 #1
    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
     
  2. jcsd
  3. Aug 26, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hihi davedave! :smile:

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

    Do you mean
    Code (Text):
    A x x x
    x x B x
    x x x C
    ?
     
  4. Aug 26, 2008 #3
    That's easy. Map out all the possible paths (from A-->B, south/east), and then count how many intersect B.
     
  5. Aug 27, 2008 #4
    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?
     
  6. Aug 28, 2008 #5
    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?
     
  7. Aug 28, 2008 #6
    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.
     
  8. Aug 28, 2008 #7
    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.
     
  9. Aug 28, 2008 #8
    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?
     
  10. Aug 29, 2008 #9
    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.
     
    Last edited: Aug 29, 2008
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