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We start with an urn that has $i$ white balls. At each step we draw $1$ ball. If it is white, we return a black ball. If it is black, we return this black ball. 1) What is the probability of drawing a white ball at the $n$-th step? - Checking the decision tree I found that $(1-1/i)^{n+1}$ works. But why? 2) How many white balls have been drawn at the $n$-th step? - Go via cumulative distribution function? Thanks :) |
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Think of all the balls being numbered from $1$ to $i$. The process described is the same as randomly changing the color of a ball to black. The probability of any ball staying white for $n$ draws is $(1-\frac{1}{i})^n$. Thus, after $n$ draws, the expected number of white balls is $i(1-\frac{1}{i})^n$ by the linearity of expectation. Since the probability of drawing a white ball is the number of white balls divided by $i$, and the probability of drawing a white ball is the expected probability, the probability of drawing a white ball is the expected number of white balls divided by $i$. Thus, the probability of drawing a white ball on the $n^{th}$ draw is $(1-\frac{1}{i})^{n-1}$. After $n$ draws the expected number of white balls is $i(1-\frac{1}{i})^n$, so the expected number of white balls drawn would be $i-i(1-\frac{1}{i})^n$. |
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