Let there be $n$ bins. Each $i^{th}$ bin contains $n-i$ white balls and $i-1$ black balls. If a bin is picked at random from which two balls are picked without replacement. What would be the probability of the second ball to be white?
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2 Answers
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If you were to renumber bin $i$, $i=1, 2, \ldots, n$ as bin $n+1-i$--which clearly cannot affect the results--your question would read
Bin $n+1-i$ contains $n-(n+1-i)=i-1$ white balls and $(n+1-i)-i=n-i$ black balls. ... What would be the probability of the second ball to be white?
Upon exchanging "black" for "white" throughout, it's exactly the same question as the original. Consequently, the probability the second ball is white must equal the probability the second ball is black. Now apply the Law of Total Probability to obtain the answer.
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Denote the probability of the event that the second ball is white as $P(white)$. You want to evaluate $P(white)=\sum_{i=1}^nP(white|bin=i)P(bin=i)$. For $P(white|bin=i)$, you want the second ball to be white. There are a total of $n-i+i-1=n-1$ balls in each bin. This is equivalent to either $WW$ or $BW$ being drawn:
$P(white|bin=i)=\frac{n-i}{n-1}\cdot \frac{n-i-1}{n-2}+\frac{i-1}{n-1}\frac{n-i}{n-2}$.
Can you finish it from here?
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$\begingroup$ I believe you can obtain the same formula, but in a simpler and more insightful form, by observing that the chance the second ball is white must equal the chance the first ball is white, which obviously is $(n-i)/(n-1)$. $\endgroup$– whuber ♦Commented Jan 22, 2016 at 21:25
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$\begingroup$ @whuber: I came up with that too for the first ball. Bear in mind that this is done without replacement. Besides the problem seems to have a rational number as answer. Could there be a way to cancel out these terms? $\endgroup$ Commented Jan 23, 2016 at 10:20