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Math Help - probability question

  1. #1
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    probability question

    suppose that a large number of people standing in a queue are asked to reorder themselves so that they line up in order of height. what is the approximate probability that at least one person is in the same position after the reordering as they were before?

    i did up lists for up to 4 people and the results are:

    1 person, Pr = 1 = 1
    2 people, Pr = 1/2 = 0.5
    3 people, Pr = 2/3 = 0.667
    4 people, Pr = 5/8 = 0.625

    i cannot for the life of me see a pattern, considering the 2 people probability means that the probabilities aren't even in decreasing order. i have done that particular probability correct, right?

    also, the question has a hint:


    ex = S xr
    r=0 r!

    WHAT DOES THIS HAVE TO DO WITH THE QUESTION?!?!
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    suppose that a large number of people standing in a queue are asked to reorder themselves so that they line up in order of height. what is the approximate probability that at least one person is in the same position after the reordering as they were before?

    i did up lists for up to 4 people and the results are:

    1 person, Pr = 1 = 1
    2 people, Pr = 1/2 = 0.5
    3 people, Pr = 2/3 = 0.667
    4 people, Pr = 5/8 = 0.625

    i cannot for the life of me see a pattern, considering the 2 people probability means that the probabilities aren't even in decreasing order. i have done that particular probability correct, right?

    also, the question has a hint:


    ex = S xr
    r=0 r!

    WHAT DOES THIS HAVE TO DO WITH THE QUESTION?!?!
    1 - Pr(no person in same position).

    The concept of derangement is what you need to get this probabilty. Read Derangement - Wikipedia, the free encyclopedia.
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  3. #3
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    ok. so the formula for number of derangements is
    C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG

    that means that the approximate probability that at least one person is in the same position after the reordering as they were before is
    1-C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG
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  4. #4
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    whoops that came out funny ill just type it. the formula is

    n! sum of (from r=0 to n) [(-1)^2]/r!

    so the answer to the question is just 1-this formula?
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  5. #5
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    Quote Originally Posted by wik_chick88 View Post
    ok. so the formula for number of derangements is
    C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG

    that means that the approximate probability that at least one person is in the same position after the reordering as they were before is
    1-C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG
    For a large number of people the probability that no person is in the same position is 1/e.

    So the probability of at least one person in the same position is 1 - \frac{1}{e}.
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  6. #6
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    ok. i THINK i understand. its asking for the probability, so with the derangement formula, you dont need the n! at the start becuase you arent looking for a number, just a probability. then, the hint means that as r goes from 0 to infinite, the sum of (x^r)/r! equals e^x. so that means that if x = -1 such as in the derangement formula, as r goes from 0 to infinite, the sum of (-1)^r/r! actually equals e^-1 which equals 1/e. YAYY!!! thanks so much for your help!!
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