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Consider an urn with two type of balls white and black with proportionality $p,q$ respectively such that $p+q=1$ we extract balls $n$-time successively with replacement. calculate the probability that the k-th extracted white ball appears on the n-th extraction? so here is my idea since we have the probability that a white ball appears for the first time at the n-th extraction is $q^{n-1} p$ and the probability to extract exactly k white balls from the n extractions is : $\binom{n}{k} q^{n-k} p^k$ there must be a way to connect this two to find the answer of this question no?

HellBoy
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2 Answers2

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We want to have exactly $k-1$ white balls extracted in the first $n-1$ extractions, and then to extract a white ball. This probability is given by $${{n-1}\choose {k-1}}p^{k-1}q^{n-1-(k-1)} \cdot p={{n-1}\choose {k-1}}p^{k}q^{n-k}$$

HappyDay
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The approach is correct, but not quite there.

You want the probability for obtaining $k-1$ white among the first $n-1$ draws, and a white ball on the subsequent draw (and thus it will be the $k$-th white ball on the $n$-th draw).

Graham Kemp
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  • but why are we using the binomial coefficient doesnt that reflect some type of order? meanwhile the extractions are done with replacement so order doesnt really matter? – HellBoy Oct 08 '23 at 11:39
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    The binomial coefficient reflects the fact that we are counting all possible combinations of extractions in which $k-1$ balls are extracted in the first $n-1$ attempts. – HappyDay Oct 08 '23 at 11:45
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    Order does not matter, so we must count all possible orders of selection of places for the white balls. – Graham Kemp Oct 08 '23 at 11:53