I suspect this isn't optimal, but this seems like the straightforward way to me.
Mark each computer by affixing a post-it note (or a piece of gum if you really like gum).
Select at random one marked computer. Ask all marked computers (including the one selected) whether the selected computer is faulty. If a majority answer that the computer is working, you have found your working computer.
Otherwise, if a majority of computers say the selected computer is faulty, then unmark the selected computer by removing the post-it note. Also unmark one other randomly selected computer. Proceeding this way, each round should end with either the answer (yay) or two fewer marked computers. Because at least one of the two computers you eliminate on each round is known to be faulty, you are guaranteed to maintain a majority of working computers.
In the worst case you have 49 faulty computers that always answer "working." And you unluckily select a faulty computer to evaluate every round. And you unluckily randomly eliminate a working computer on every round (along with the faulty computer discovered).
In the last (49th) round of this worst case scenario, you have three computers of which one is faulty and two are working, so once you determine the faulty one, both the remaining computers are working and you have your answer.
In this worst case, you will ask 99 + 97 + 95 + ... + 3 = 2,499 questions.
This worst case is pretty bad, but typically you would have to be quite unlucky to go more than a few rounds before randomly choosing a working computer to evaluate. The average case is probably more like 200 questions.
Edit: Hmm, should have thought more about this one before answering. As the above answers (or googling for "engineers and managers puzzle") demonstrate, this approach is indeed embarrassingly far from optimal... Oops.
For anyone interested in this question that is ready for a spoiler, there is an excellent discussion about how to use a tree to do considerably better than N-2 for many values of N at