Distinguishable or Indistinguishable: Occupancy Problem Problem Statement: Contracts for two construction jobs are randomly assigned to one or more of three firms, $A, B,$ and $C.$ Let $Y_A$ denote the number of contracts assigned to firm $A,$ and $Y_B$ the number of contracts assigned to firm $B.$ Recall that each firm can receive $0,1,$ or $2$ contracts. Find the joint probability function for $Y_A$ and $Y_B.$
My Work So Far: To begin analyzing the situation, this is an occupancy problem, as balls in urns. The contracts
are being treated as indistinguishable, whereas the firms clearly are. As it is possible for a firm
not to get a contract, it follows that the number of ways to assign the contracts is
$$\binom{3+2-1}{2}=\binom42=6.$$
These correspond to the following options:
$$ 
\begin{array}{ccc}
Y_A &Y_B &Y_C\\ \hline
2 &0 &0\\
1 &1 &0\\
1 &0 &1\\
0 &2 &0\\
0 &1 &1\\
0 &0 &2
\end{array}
$$
Each is equi-probable, hence the joint probability function for $Y_A$ and $Y_B$ is $1/6$ for each row. If we line up the variables for the joint distribution, we obtain:
$$ 
\begin{array}{c|ccc}
Y_B\downarrow Y_A\to&0 &1 &2\\ \hline
0 &1/6 &1/6 &1/6\\
1 &1/6 &1/6 &0\\
2 &1/6 &0 &0\\
\end{array}
$$

Here is my question: how do I know whether the problem is talking about distinguishable contracts, or indistinguishable contracts? I answered that the contracts are indistinguishable, since the variable of interest appears to be just the total number of contracts. The book's answer clearly assumes distinguishable contracts:

$$ 
\begin{array}{c|ccc}
Y_B\downarrow Y_A\to&0 &1 &2\\ \hline
0 &1/9 &2/9 &1/9\\
1 &2/9 &2/9 &0\\
2 &1/9 &0 &0\\
\end{array}
$$

I have a very nice chart of occupancy problem solutions, which I won't bother to replicate here; suffice it to say that the number of ways to put balls into urns varies greatly depending on three variables: whether the balls are distinguishable, whether the urns are distinguishable, and whether urns are allowed to be empty.  The chart says that if the contracts are distinguishable, the firms are distinguishable, and a firm can get zero contracts, then there are $3^2=9$ ways to award the contracts, which makes sense.
How do I tell which case I'm in? Do I just infer from the nature of contracts that one contract couldn't be the same as another? I could imagine contracts (like manufacturing contracts, e.g.) that would not be distinguishable.
Note that M.SE has many, many questions regarding how to calculate the right answer, given that you know the balls and urns are distinguishable or indistinguishable. My question is one step back from that: given a (in my opinion) poorly worded question that doesn't specify whether the contracts are distinguishable or not, how do I tell the difference?

Thanks very much for your time!
 A: The number of possible combinations depend on (in)distinguishability of the objects of concern, however the odds doesn't. The observer not being able to distinguish objects doesn't change the probability of events. So, in probability calculations it doesn't matter if some object is distinguishable or not, and the good news is you won't even try to infer it from the question.
Edit:
While calculating probabilities, you cannot divide "number of all different situations" by "number the situations you are interested in" when all the situations are not equally likely. Each situation has a different probability mass. For example, in your calculations, you say that $_=2$ has $1/6$ probability. However, probability of $$ getting both contracts is simply $1/3×1/3$. $(2,0,0)$ and $(1,1,0)$ cases don't have the same mass!
A simpler example: take two different urns, and two indistinguishable balls. The number of ways we could put these balls into these urns is $3$: (0,2),(1,1),(2,0). On the other hand, the probability of the each urn having one ball (i.e. the case (1,1)) is not 1/3. It's 1/2. 
Distinguishable or not, each urn having one ball could have happened two ways: first ball goes into urn 1, second ball goes into urn 2; or the other way around. (0,2) case can happen only one way: first and second ball goes into Urn 2. Similarly, (2,0) case can happen one way, adding up to a total of 4 cases, in which two of them are of interest two us, which yields $2/4=1/2$.
