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Math Help - Conditinal Probability with numbers selected at random

  1. #1
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    Conditinal Probability with numbers selected at random

    Suppose that three numbers are selected one by one, at random, and without replacement from the set of numbers {1, 2, 3, ... n}. What is the probability that the third number falls between the first two if the first is smaller than the second?

    The book tells me that the answer is 1/3, but that isn't much help towards how to find the answer.
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  2. #2
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    Hello, Zennie!

    \text{Suppose that three numbers are selected randomly without replacement}
    \text{from the set of numbers }\{1,\,2,\,3,\,\hdots\, n\}.

    \text{What is the probability that the third number falls between the first two}
    \text{if the first is smaller than the second?}

    \text{The book answer is: }\frac{1}{3}

    Let the three numbers be: . a,\,b,\,c

    There are 3! = 6 permutations of the three numbers:
    . . abc,\:acb,\:bac,\:bca,\:cab,\:cba . (in increasing order)


    "The first is less than the second": .  a < b
    . . There are three cases: . abc,\:acb,\:cab

    In only one case, \,c is between \,a and \,b: . acb

    Therefore, the probability is: . \dfrac{1}{3}
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  3. #3
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    Wow, thanks a lot. That's so much more simple than I was expecting. I probably wouldn't have thought about it that way.
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  4. #4
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    It is not such a tough question. Try using this hint:
    Pick three number (in order) - x1,x2,x3 from {1,2,3,.....,n}
    What is the Prob(x1<x2<x3)? [You don't need any calculation - think symmetry !!]
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