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klimenkov
 3 years ago
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.
klimenkov
 3 years ago
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.

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amriju
 3 years ago
Best ResponseYou've already chosen the best response.0is the answer: 1/2{\[\sum_{1}^{n} (1/r) \]}

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Hm.. Please, write it using "Draw" button.

amriju
 3 years ago
Best ResponseYou've already chosen the best response.0is this the answer...? or does it have a definite answer?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0It has a definite answer and your answer is not right.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1let 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 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am confused as to why it is not one half by pure reason replace the word "white" with "black" and nothing has changed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ya looks like either 1 or 0.

klimenkov
 3 years ago
Best 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0one 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
 3 years ago
Best ResponseYou've already chosen the best response.0lets 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
 3 years ago
Best ResponseYou've already chosen the best response.0actually it is kind of cute isn't is? you get \[\frac{1}{n(n+1)}\sum_{k=1}^nk\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how is that for a "strict" proof?
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