A confusing question on probability 
In a race, the probabilities of A and B winning the race are
  $\frac{1}{3}$  and $\frac{1}{6}$  respectively. Find the probability
  of neither of them winning the race.

I solved the question in the following manner-
Since A and B are running in a race, probability of neither of them winning is $$1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}$$
However, all the solution books that I refer to are solving it in the following manner
$$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)=\frac{5}{9}$$
Now this does not make any sense to me since the events of A winning and B winning are not independent. I was pretty confident of my answer but even the official answer key of the test has given $\frac{5}{9}$ as the answer. 
Where exactly am I wrong? How can the winning of A and B be independent of each other since given that one does not win, the winning chances of the other increases?
 A: As has been discussed in the comments, your interpretation of the question makes much more sense than one that might lead to these events being viewed as independent. Even if this were a race that can be won by more than one participant, it would be a strange race indeed if the chance of winning it were independent of someone else’s chance of winning it.
I did find the wrong answer that you found online, but your correct answer is also online here.
A: 
Important
Since $A$ and $B$ are not opponents (this race is not zero-sum),then the probabilities of winnings do not sum up to $1$ and it is possible for both $A$ and $B$ to win the race simultaneously.

Your mistake in summing up the probabilities is that you have considered $$1-[P(A)+P(B)]$$instead of $$P(A'\cap B')=1-[P(A\cup B)]$$i.e. you have assumed that it is impossible for both $A$ and $B$ to win the race.
Remark
The case of not winning one of them, does not increase the chance for the other one's winning. You can equivalently assume that neither of them is aware of the state of the other one and the race is simply a math competition during limited time.
