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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.

  • one year ago
  • one year ago

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  1. amriju
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    is the answer: 1/2{\[\sum_{1}^{n} (1/r) \]}

    • one year ago
  2. klimenkov
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    Hm.. Please, write it using "Draw" button.

    • one year ago
  3. amriju
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    |dw:1350863939368:dw|

    • one year ago
  4. amriju
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    is this the answer...? or does it have a definite answer?

    • one year ago
  5. klimenkov
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    It has a definite answer and your answer is not right.

    • one year ago
  6. hartnn
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    *

    • one year ago
  7. shubhamsrg
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    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 ?

    • one year ago
  8. satellite73
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    i am confused as to why it is not one half by pure reason replace the word "white" with "black" and nothing has changed

    • one year ago
  9. sauravshakya
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    Ya looks like either 1 or 0.

    • one year ago
  10. klimenkov
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    @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!

    • one year ago
  11. xxjakkixx
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    1 or 0

    • one year ago
  12. satellite73
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    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

    • one year ago
  13. satellite73
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    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 \(n+1\)

    • one year ago
  14. satellite73
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    actually it is kind of cute isn't is? you get \[\frac{1}{n(n+1)}\sum_{k=1}^nk\]

    • one year ago
  15. satellite73
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    how is that for a "strict" proof?

    • one year ago
  16. klimenkov
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    Very nice!

    • one year ago
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