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.

Question

amriju · Answer

is the answer: 1/2{$$\sum_{1}^{n} (1/r) $$}

klimenkov · Answer

Hm.. Please, write it using "Draw" button.

amriju · Answer

|dw:1350863939368:dw|

amriju · Answer

is this the answer...?
or does it have a definite answer?

klimenkov · Answer

It has a definite answer and your answer is not right.

hartnn · Answer

*

shubhamsrg · Answer

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 · Answer

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 · Answer

Ya looks like either 1 or 0.

klimenkov · Answer

@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 · Answer

1 or 0

anonymous · Answer

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 · Answer

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 · Answer

actually it is kind of cute isn't is?  you get 
$$\frac{1}{n(n+1)}\sum_{k=1}^nk$$

anonymous · Answer

how is that for a "strict" proof?

klimenkov · Answer

Very nice!