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klimenkov
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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
klimenkov Group Title
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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amriju Group TitleBest ResponseYou've already chosen the best response.0
is the answer: 1/2{\[\sum_{1}^{n} (1/r) \]}
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Hm.. Please, write it using "Draw" button.
 one year ago

amriju Group TitleBest ResponseYou've already chosen the best response.0
dw:1350863939368:dw
 one year ago

amriju Group TitleBest ResponseYou've already chosen the best response.0
is this the answer...? or does it have a definite answer?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
It has a definite answer and your answer is not right.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
let x be no. of white balls, then no. of black balls = nx probab of taking 1 white ball = x/n of taking 2 white balls = (x/n)* ( x1 / n1) of 3 balls = (x/n) * (x1 /n1) * ( x2 / n2) . . . given that all are equal x/n = x/n * (x1)/(n1) = (x/n) * (x1)/(n1) * (x2)/(n2) ..... =>x=n ans comes out to be 1 !! where did i go wrong ?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Ya looks like either 1 or 0.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
@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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
actually it is kind of cute isn't is? you get \[\frac{1}{n(n+1)}\sum_{k=1}^nk\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
how is that for a "strict" proof?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Very nice!
 one year ago
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