klimenkov
  • klimenkov
There are \(n\) balls in the box. There are only black or white balls in the box. The number of white balls is unknown. Assume that that any number of white balls is equiprobable and find a probability to take a white ball from the box if the ball is taken from the box randomly.
Probability
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SOLVED
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katieb
  • katieb
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amriju
  • amriju
is the answer: 1/2{\[\sum_{1}^{n} (1/r) \]}
klimenkov
  • klimenkov
Hm.. Please, write it using "Draw" button.
amriju
  • amriju
|dw:1350863939368:dw|

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More answers

amriju
  • amriju
is this the answer...? or does it have a definite answer?
klimenkov
  • klimenkov
It has a definite answer and your answer is not right.
hartnn
  • hartnn
*
shubhamsrg
  • shubhamsrg
let x be no. of white balls, then no. of black balls = n-x probab of taking 1 white ball = x/n of taking 2 white balls = (x/n)* ( x-1 / n-1) of 3 balls = (x/n) * (x-1 /n-1) * ( x-2 / n-2) . . . given that all are equal x/n = x/n * (x-1)/(n-1) = (x/n) * (x-1)/(n-1) * (x-2)/(n-2) ..... =>x=n ans comes out to be 1 !! where did i go wrong ?
anonymous
  • anonymous
i am confused as to why it is not one half by pure reason replace the word "white" with "black" and nothing has changed
anonymous
  • anonymous
Ya looks like either 1 or 0.
klimenkov
  • klimenkov
@satellite73 Can you prove it strictly? @shubhamsrg Why do you find a probability of taking 2,3,.. white balls? We just need to find a probability of taking just 1 white ball!
anonymous
  • anonymous
1 or 0
anonymous
  • anonymous
one is a silly answer, that means you are sure to pick a white ball likewise so is zero, that means you never get one you cannot tell black from white in this problem, therefore, by symmetry, it is \(\frac{1}{2}\) unless for some reason i am reading the problem wrong
anonymous
  • anonymous
lets say there were four balls in the box, so there are five equally likely number of white balls \(\{0,1,2,3,4\}\) each of which occur with probability \(\frac{1}{5}\) then the probability you get a white ball is \[0\times \frac{1}{5}+\frac{1}{4}\times \frac{1}{5}+\frac{2}{4}\times \frac{1}{5}+\frac{3}{4}\times \frac{1}{5}+\frac{4}{4}\times \frac{1}{5}\] \[=\frac{1+2+3+4}{20}=\frac{1}{2}\] now if you want some sort of formula argument, replace \(4\) by \(n\) and \(5\) by \(n+1\)
anonymous
  • anonymous
actually it is kind of cute isn't is? you get \[\frac{1}{n(n+1)}\sum_{k=1}^nk\]
anonymous
  • anonymous
how is that for a "strict" proof?
klimenkov
  • klimenkov
Very nice!

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