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Math Help - Work out this probability

  1. #1
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    Exclamation Work out this probability

    "Write an expression for the probability that in a sequence of r random integers in the range [1,n] no integer occurs more than once"

    Is my answer right?

    Let the sequence be represented by a_{n}= a_{1}, a_{2}, ... a_{r} where r<=n

    P(a_{1}=a_{2}) = \frac{1}{n}
    P(a_{1}!=a_{2}) = 1-\frac{1}{n}

    So P(a_{1}!=a_{2}!=...!=a_{r}) = (1-\frac{1}{n})^r
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  2. #2
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    Quote Originally Posted by DaRush19 View Post
    "Write an expression for the probability that in a sequence of r random integers in the range [1,n] no integer occurs more than once"

    Is my answer right? No it is not.
    Let the sequence be represented by a_{n}= a_{1}, a_{2}, ... a_{r} where r<=n
    P(a_{1}=a_{2}) = \frac{1}{n}
    P(a_{1}!=a_{2}) = 1-\frac{1}{n}
    So P(a_{1}!=a_{2}!=...!=a_{r}) = (1-\frac{1}{n})^r
    Here is a hint.
    Roll a die three times. The probability that no number occurs more than once is \frac{6\cdot 5\cdot 4 }{6^3}.
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  3. #3
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    Now that I've worked it out as (n(n-1)(n-2)..(n-r+1))/n^r I'm stuck on when this is approximately equal to 1/2

    :S
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