Probability Question 3.17 in Wackerly, Mendenhall, and Scheaffer, 5th Ed. 
Problem 3.17. Two construction contracts are to be randomly assigned to one or more of three firms:
  I, II, and III. Any firm may receive both contracts. If each contract will yield a profit of
  $\$90,000$ for the firm, find the expected profit for firm I. If firms I and II are actually owned
  by the same individual, what is the owner's expected total profit?

My Answer. Let $Y$ be the number of contracts assigned to firm I. Then this problem is equivalent to
indistinguishable balls put into distinguishable urns, where the urns can be empty. That means
the number of ways to assign $2$ indistinguishable contracts to $3$ distinguishable firms is
$$\binom{2+3-1}{2}=\binom{4}{2}=6. $$
It follows that the distribution for $Y$ is as follows:
$$
\begin{array}{c|c|c|c}
y      &0    &1    &2    \\ \hline
P(Y=y) &1/2  &1/3  &1/6
\end{array}
$$
This means that the expected profit, given by $E(\$90000Y),$
is
$$E(\$90000Y)=\$90000\left[0/2+1/3+2/6\right]=\$60000. $$
Now if one owner owns two firms, let $X$ be a random variable describing how many contracts either
company I or II got. This essentially coalesces the two companies I and II in terms of our analysis. 
We have the following probability distribution for $X:$
$$
\begin{array}{c|c|c|c}
x      &0    &1    &2    \\ \hline
P(X=x) &1/3  &1/3  &1/3
\end{array}
$$
It follows that the expected value for the total profit is given by
$$E(\$90000X)=\$90000[0/3+1/3+2/3]=\$90000. $$

My Question. The answer in the back of the book, however, is $\$120000.$ Why is that? Is there an error in my reasoning? Thanks in advance for your time!


There's a problem with your reasoning: each combination is not just as likely to occur. If each contract has a probability of $\frac{1}{3}$ to go to one of the firms, than the probabilities are as follows:
$$P(0) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(1) = \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(2) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$$
We find:
$$E[X] = \frac{4}{9} \cdot 0 + \frac{4}{9} \cdot 90000 + \frac{1}{9} \cdot 180000 = 60000$$
You could also arrive here by dividing the total expected profit for all companies by the number of companies, or $180000$ divided by three. If one individual owns both companies, we can simply multiply by two and arrive at $120000$. Alternatively:
$$P(0) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$$
$$P(1) = \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(2) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
We then find:
$$E[X] = \frac{1}{9} \cdot 0 + \frac{4}{9} \cdot 90000 + \frac{4}{9} \cdot 180000 = 120000$$
