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''
It is easier to work out the chance of not picking out the different sheet and subtract it from 1.
Originally Posted by Robertf1
If x=2 then your chance of not getting the different sheet is and so the chance of success is which is 75%.
Try it for x=3 and then the general case