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.

Re: Question about probability for things ''without memory''

Quote:

Originally Posted by

**Robertf1** 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 $\displaystyle \left(\dfrac{1}{2}\right)^2$ and so the chance of success is $\displaystyle 1 - \left(\dfrac{1}{2}\right)^2$ which is 75%.

Try it for x=3 and then the general case