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Math Help - Some probability

  1. #1
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    Some probability

    A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?



    Moderator Edit: Approved Challenge question.
    Last edited by mr fantastic; April 19th 2010 at 10:11 PM.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?



    Moderator Edit: Approved Challenge question.
    The length of the segment is irrelevant (provided it is nonzero).

    Spoiler:

    The distinction between > and \geq doesn't matter here.

    WLOG let the segments be a, b, c with a \leq b \leq c.

    Triangle inequality: a+b > c

    In fact this is equivalent to c < \frac{1}{2}, for the given constraints. ( a+b+c=1 so we can substitute using a+b=1-c.)

    (Edited out a false start.)

    So we have two variables x, y with uniform probability distribution in [0,1]. I made a sketch:



    The shaded area is the probability we want, which is p = \frac{2}{8} = \frac{1}{4}.

    Last edited by undefined; April 20th 2010 at 10:16 AM.
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    Quote Originally Posted by undefined View Post
    The length of the segment is irrelevant (provided it is nonzero).

    Spoiler:

    The distinction between > and \geq doesn't matter here.

    WLOG let the segments be a, b, c with a \leq b \leq c.

    Triangle inequality: a+b > c

    In fact this is equivalent to c < \frac{1}{2}, for the given constraints. ( a+b+c=1 so we can substitute using a+b=1-c.)

    (Edited out a false start.)

    So we have two variables x, y with uniform probability distribution in [0,1]. I made a sketch:



    The shaded area is the probability we want, which is p = \frac{2}{8} = \frac{1}{4}.

    Good job.
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  4. #4
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    I let  x+y<1
    ,  x+y > \frac{1}{2}
     x,y < \frac{1}{2}

    The area of the region bounded is also one fourth of the triangle .
    Last edited by simplependulum; April 21st 2010 at 12:28 AM.
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