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A box contains m white balls and n black balls. We extract them one by one, without reintroducing them back. What is the probability to obtain the first ball white at the k-th extraction?

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In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is $$\Pr\lbrack \mbox{first $k-1$ balls black} \rbrack =\prod_{i=0}^{k-2} \frac{n-i}{m+n-i} $$

The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then

$$\Pr\lbrack \mbox{$k^{th}$ ball is white | first $k-1$ balls black} \rbrack =\frac{m}{m+n-(k-1)} $$

The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.

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  • $\begingroup$ I understand, thank you for your answer! $\endgroup$ Commented Feb 1, 2019 at 21:33

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