a) We have 3 initially empty urns. One of them is chosen at random with equal probability (1/3 for each). Then, a ball is placed inside the chosen urn. The process is repeated until one of the urns has two balls. What is the probability that when the process ends, the total number of balls in all urns is equal to 2?

b) Now we have 2014 initially empty urns. Repeating the same process as before, what is the probability that at the end there will be exactly 10 balls in the urns?
Solution

a) First, a ball is placed in some urn. For the process to end with two balls, it is necessary that, in the next step, the next ball is placed in the same urn where the first ball was placed. This has a probability of 1/3. Therefore, the probability that the process ends with two balls is equal to $1 / 3$.

b) First, a ball is placed in some urn. In the next step, it is necessary that the next ball be placed in a different urn from the first one chosen. Otherwise, the process ends! Since there are 2013 empty urns (one is already occupied), the probability of this happening is

$$
\frac{2013}{2014}
$$

In the third step, a new urn is chosen. For the process not to end, it is necessary that this new urn chosen is different from the previous two. The probability of this happening is

$$
\frac{2012}{2014}
$$

In the fourth step, a new urn is chosen. By the same argument as before, the probability that the chosen urn is different from the previous ones is

$$
\frac{2011}{2014}
$$

and so on until the ninth urn, whose probability will be

$$
\frac{2006}{2014}
$$

For the process to end with ten balls, it is necessary that the tenth ball be placed in an urn that already has a ball! Since we have, up to this point, nine occupied urns and there are 2014 urns, the probability of this occurring is

$$
\frac{9}{2014}
$$

Thus, the probability that the process ends with 10 balls is equal to the product of the probabilities described above, which gives us

$$
\frac{2013}{2014} \cdot \frac{2012}{2014} \cdot \frac{2011}{2014} \cdot \frac{2010}{2014} \cdot \frac{2009}{2014} \cdot \frac{2008}{2014} \cdot \frac{2007}{2014} \cdot \frac{2006}{2014} \cdot \frac{9}{2014}
$$

as the result.