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Tafe is an abbreviation. From dictionary.com:

 

 

TAFE: Transverse Alternating-Field Electrophoresis

 

 

Now you know.

A man has n keys of which only one opens his office door. He tries the keys in random order, possible trying the same key several times. Find P®, the probability that he opens the door on the Rth attempt and show that the infinite sum Pr = 1.
SupYouFool;550119']A man has n keys of which only one opens his office door. He tries the keys in random order' date=' possible trying the same key several times. Find P®, the probability that he opens the door on the Rth attempt and show that the infinite sum Pr = 1.[/quote']

 

P(1) is 1/n times <chance he didn't get it open on first try, which is (1-P(1))

= 1/n * (1 - 1/n) = 1/n - 1/n^2

 

And P(3) becomes 1/n * (1 - (1/n - 1/n^2))) = 1/n - 1/n^2 + 1/n^3

 

So P(i) = Sum<j=0-i> (-1/n^i) Which is a simple powers series.

 

and yes, this is my work...you can't find a problem like that in google.

P(1) is 1/n times <chance he didn't get it open on first try, which is (1-P(1))

= 1/n * (1 - 1/n) = 1/n - 1/n^2

 

And P(3) becomes 1/n * (1 - (1/n - 1/n^2))) = 1/n - 1/n^2 + 1/n^3

 

So P(i) = Sum<j=0-i> (-1/n^i) Which is a simple powers series.

 

and yes, this is my work...you can't find a problem like that in google.

 

Cool, I fucking hate this class : ( I cannot do stats high, and therefore I can't do stats.

P(1) is 1/n times <chance he didn't get it open on first try, which is (1-P(1))

= 1/n * (1 - 1/n) = 1/n - 1/n^2

 

And P(3) becomes 1/n * (1 - (1/n - 1/n^2))) = 1/n - 1/n^2 + 1/n^3

 

So P(i) = Sum<j=0-i> (-1/n^i) Which is a simple powers series.

 

and yes, this is my work...you can't find a problem like that in google.

 

Hmmmmm... Seven ?

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