## 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

1. amriju

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

2. klimenkov

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

3. amriju

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4. amriju

5. klimenkov

6. hartnn

*

7. 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 ?

8. satellite73

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

9. sauravshakya

Ya looks like either 1 or 0.

10. 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!

11. xxjakkixx

1 or 0

12. satellite73

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

13. satellite73

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 $$n+1$$

14. satellite73

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

15. satellite73

how is that for a "strict" proof?

16. klimenkov

Very nice!