a generic ball and urn problem 
Generic ball and urn problem: If two balls are randomly placed in three urns, what are the chances the second urn is occupied?

In the generic ball and urn problem, since the balls aren't stated to be distinct why can't I solve this problem by saying there are six possible ways of arranging the 2 balls in 3 urns i.e.
sample space: ${(b|b| )\;, \;(b| |b)\;, \;( |b|b)\;, \;(bb| | )\;, \;( |bb| )\;, \;( | |bb)}$
and 3 possible ways of filling the second urn and so the probability should be $\frac{3}{6}$ instead of $\frac{5}{9}$.
Why is this not correct? This has been bugging me from quiet some time. Could someone help?
 A: We may assume that the balls are distinct in order to allow for our sample space to be equiprobable.  In the case that the balls are indistinct, the scenario $\underbrace{(~~)}_{\text{urn 1}}~\underbrace{(bb)}_{\text{urn 2}}~\underbrace{(~~)}_{\text{urn 3}}$ is half as likely as $\underbrace{(b~)}_{\text{urn 1}}~\underbrace{(b~)}_{\text{urn 2}}~\underbrace{(~~)}_{\text{urn 3}}$
Assuming that the balls are distinct then, we may approach this via direct counting, via indirect counting, or via principles of probability.
Let the balls be colored red and blue.  Our sample space is then: $\{(\color{red}{b}\color{blue}{b})(~~)(~~)\},\{(\color{red}{b})(\color{blue}{b}~)(~~)\},\{(\color{red}{b})(~~)(\color{blue}{b}~)\}, \{(\color{blue}{b}~)(\color{red}{b}~)(~~)\},\{(~~)(\color{red}{b}\color{blue}{b})(~~)\},\dots,\{(~~)(~~)(\color{red}{b}\color{blue}{b})\}$
There are $3^2=9$ elements in our sample space (Which urn did the red ball go to?  Which urn did the blue ball go to?) and each of these are equally probable.  We need the outcomes to be equally probable to use the definition of probability that allows $Pr(A)=\frac{\text{# of outcomes in}~A}{\text{# of outcomes total}}$.
By brute force counting, you see there are $5$ possibilities where the second urn is occupied.
Alternatively, approaching via inclusion-exclusion, we see that to have the second urn unoccupied, the red and blue balls each have only two choices for where they can go, for a total of $2^2=4$ ways to have the second urn unoccupied.  Subtracting from the total again gives $5$ possibilities where it is occupied.
In either scenario, we divide by the number of cases to get the probability is $\frac{5}{9}$.

Using probability theory instead, consider approaching via multiplication principle and calculating the probability of the opposite event.
There is a $\frac{2}{3}$ chance that the first ball is not placed in the second urn.  Again, there is a $\frac{2}{3}$ chance the second ball is not placed in the second urn.  Since these are assumed to be independent events, by multiplying, we have $\frac{4}{9}$ as the probability that the second urn is unoccupied.  Subtracting from one, we have $1-\frac{4}{9}=\frac{5}{9}$ is the probability that the second urn is occupied.
A: Very often, the elements of a sample space will have unequal probabilities.
Your sample space is an example.
In the usual way of putting balls into urns, we put one ball into an urn at random, then we do this again with a second ball. At that time one of the three urns is already occupied, and the second ball has a $\frac13$ probability
of being put in that urn, $\frac23$ probability of being put in one of the
other two urns.
So unless you have some unusual constraints on how you put balls into urns
that prevents exactly half of the placements of the second ball into empty urns,
the probabilities of $b|b|$, $b||b$, and $|b|b$ are each equal to $\frac29$,
and the probabilities of $bb||$, $|bb|$, and $||bb$ are each equal to $\frac19$.

It is possible to contrive a procedure that will leave the second urn
unoccupied with probability $\frac12$.
First, choose an urn at random, so each urn is chosen with probability $\frac13$.
Next, toss a fair coin. If the outcome is heads, put both balls in the urn you selected in the first step. Otherwise, set that urn aside and put one ball in each of the other two urns.
This is a random process, at the end of which each urn has a $\frac12$
chance of being empty. It is not, however, what we usually mean by
"place two balls randomly in three urns."
The usual interpretation of "place $k$ balls in $n$ urns" is that we take the balls one at a time, and each ball has an equal chance to go into each urn
regardless of what balls are in the urn already.
If someone is surreptitiously putting unique marks on each ball that you
do not notice, this should not affect where the balls go
(assuming the balls and urns are not some objects so tiny that 
quantum-mechanical effects become significant).
Hence the probability that you calculate, without accounting for labels on the balls,
should be the same as the probability that the other person calculates
using the marks that they placed.
