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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.
is the answer: 1/2{\[\sum_{1}^{n} (1/r) \]}
Hm.. Please, write it using "Draw" button.
is this the answer...? or does it have a definite answer?
It has a definite answer and your answer is not right.
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 ?
i am confused as to why it is not one half by pure reason replace the word "white" with "black" and nothing has changed
Ya looks like either 1 or 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 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
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 \(n+1\)
actually it is kind of cute isn't is? you get \[\frac{1}{n(n+1)}\sum_{k=1}^nk\]
how is that for a "strict" proof?