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Math Help - Question about probability for things ''without memory''

  1. #1
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    Question about probability for things ''without memory''

    Imagine i have a box or anything which is full with x sheets of paper. 1 sheet is different from others. I have x attempts to take blindly one sheet. What is the formula after which i can know what is the percentage of getting 'the different sheet' at least once? (If i have taken out a sheet i must put it back before next attempt!).
    As I figured out if
    x=2 then 1/2+1/2*1/2=75%
    x=3 then 1/3+1/3*1/3+1/3*1/3*1/3 (right?)

    So it means 1/x+1/x*1/x+...+(1/x)^x, but how to write it shorter, cause obviously if x=37 its big work to get result.
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  2. #2
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    Re: Question about probability for things ''without memory''

    Quote Originally Posted by Robertf1 View Post
    Imagine i have a box or anything which is full with x sheets of paper. 1 sheet is different from others. I have x attempts to take blindly one sheet. What is the formula after which i can know what is the percentage of getting 'the different sheet' at least once? (If i have taken out a sheet i must put it back before next attempt!).
    As I figured out if
    x=2 then 1/2+1/2*1/2=75%
    x=3 then 1/3+1/3*1/3+1/3*1/3*1/3 (right?)

    So it means 1/x+1/x*1/x+...+(1/x)^x, but how to write it shorter, cause obviously if x=37 its big work to get result.
    It is easier to work out the chance of not picking out the different sheet and subtract it from 1.

    If x=2 then your chance of not getting the different sheet is \left(\dfrac{1}{2}\right)^2 and so the chance of success is 1 - \left(\dfrac{1}{2}\right)^2 which is 75%.

    Try it for x=3 and then the general case
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