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Math Help - Hypergeometric/Paslow's Pyramid Question?

  1. #1
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    Hypergeometric/Paslow's Pyramid Question?

    I have no clue how to solve this question. The answer is 63/256 but I have no clue how the book got it.

    Q: You start at a corner 5 blocks south and five blocks west of your friend. You walk north and east while your friend walks south and west at the same speed. What is the probability that the two of you will meet on your travels?

    Could someone please show/explain to how to approach this?
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    Behold, the power of SARDINES!
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    Quote Originally Posted by MATHDUDE2 View Post
    I have no clue how to solve this question. The answer is 63/256 but I have no clue how the book got it.

    Q: You start at a corner 5 blocks south and five blocks west of your friend. You walk north and east while your friend walks south and west at the same speed. What is the probability that the two of you will meet on your travels?

    Could someone please show/explain to how to approach this?
    If you set up a grid with (0,0) in the bottom left corner( where you start) and (5,5) where your friend starts you should be able to convince yourself that you can only meet at the coordinates
    (0,5),(1,4),(2,3),(3,2),(4,1),(5,0)

    Now you just need to count the number of ways that you can get to each of these points.

    Note the number of paths will be symmetric for points of the form (x,y) (y,x) i.e (0,5) will be the same of (5,0).

    So there is one way to you and one way for you frend to get to (0,5)


    There are \binom{5}{1} ways for you and \binom{5}{1} for you friend to get to (1,4)

    There are \binom{5}{2} ways for you and \binom{5}{2} for you friend to get to (2,3)

    Because of the symmetry we can add all of these up to get

    2\left( \binom{5}{0}\binom{5}{0} + \binom{5}{1}\binom{5}{1} +\binom{5}{2}\binom{5}{2}\right)=2(1^2+5^2+10^2)=2  52

    There are 2^{10}=1024 paths total

    for \frac{252}{1024}
    Last edited by TheEmptySet; November 8th 2009 at 02:30 PM. Reason: Latex errors
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    Quote Originally Posted by TheEmptySet View Post
    If you set up a grid with (0,0) in the bottom left corner( where you start) and (5,5) where your friend starts you should be able to convince yourself that you can only meet at the coordinates
    (0,5),(1,4),(2,3),(3,2),(4,1),(5,0)

    Now you just need to count the number of ways that you can get to each of these points.

    Note the number of paths will be symmetric for points of the form (x,y) (y,x) i.e (0,5) will be the same of (5,0).

    So there is one way to you and one way for you frend to get to (0,5)


    There are \binom{5}{1} ways for you and binom{5}{1} for you friend to get to (1,4)

    There are \binom{5}{2} ways for you and binom{5}{2} for you friend to get to (2,3)

    Because of the symmetry we can add all of these up to get

    2\left( \binom{5}{0}\binom{5}{0} + \binom{5}{1}\binom{5}{1} +\binom{5}{2}\binom{5}{2}\right)=2(1^2+5^2+10^2)=2  52

    There are 2^10=1024 paths total

    for \frac{252}{1024}
    Thanks! I think I have the concept down. But how do we know that there are 1024 total ways? Where does the 10 come from in the 2^10?
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    Behold, the power of SARDINES!
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    Quote Originally Posted by MATHDUDE2 View Post
    Thanks! I think I have the concept down. But how do we know that there are 1024 total ways? Where does the 10 come from in the 2^10?
    Each of person on the board needs to move 5 times and there are two choices for each move.

    So there are a total of 2^5 choices for you and
    2^5 choices for your friend. So the total is

    2^5\cdot 2^5=2^{10}=1024
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