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Three urns $U_1, U_2$ and $ U_3$ each contain $5$ black balls and $7$ white balls initially. A ball is drawn at random from $U_1$ and $2$ balls of the drawn colour are added to $U_2$. Then a ball is drawn at random from $U_2$ and $3$ balls of the drawn colour are added to $U_3$. Find the probability of drawing a white ball from $ U_3$.

How do I approach these kind of problems where one event effects the other, which in turn effects some other?

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    $\begingroup$ For these kind of problems, I find trees are a good way to approach them. $\endgroup$ – Kolmin Apr 30 '16 at 19:45
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$U=(N_{black},N_{white})$

At the start :

$U_1=U_2=U_3=(5,7)$

The probability of picking a random black ball from $U_1$ is $\frac{5}{12}$, the probability of picking a white one is $\frac{7}{12}$

So once the first step (picking a ball from $U_1$) is completed, we have :

$U_2'=(7,7)$ with probability $\frac{5}{12}$

$U_2''=(5,9)$ with probability $\frac{7}{12}$

So we can pretend we are handling $\bar U_2=\frac{5}{12}U_2'+\frac{7}{12}U_2''=(\frac{35}{12},\frac{49}{12})$ probability-wise.

This is just a trick to make working with probabilities easier. The balls are still intact, and not actually cut in several twelvthes.

We follow the same process with $U_2$. The probability of picking black is $\frac{35}{84}$ while the probability of picking white is $\frac{49}{84}$.

So we get

$U_3'=(8,7)$ with probability $\frac{35}{84}$

$U_3''=(5,10)$ with probability $\frac{49}{84}$

Taking the mean, we have $\bar U_3=(\frac{25}{8},\frac{35}{8})$

So ultimately, the probability of picking white is $\frac{35}{60}=\frac{7}{12}$.

(As suggested in a comment, drawing a tree would certainly make the problem much easier to grasp)

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After the first transfer, there will be $14$ balls in $U_2$, and after the second one, $15$ balls in $U_3$

Using the concept of expectation, and focussing on white balls for shortening computations,

expected # of white balls in $U_2$ after first draw = $7+\frac7{12}\cdot2 = \frac{49}{6}$

and in $U_3$ after second draw = $7 + \frac{49/6}{14}\cdot3 = \frac{35}4$

$Pr =\frac{35/4}{15} = \frac{7}{12}$

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