Bunuel wrote:
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
A. \(\frac{1}{33}\)
B. \(\frac{2}{33}\)
C. \(\frac{1}{3}\)
D. \(\frac{16}{33}\)
E. \(\frac{11}{12}\)
Still another approach: I am surprised no one has yet posted a solution that adopts a less rigorous, more logic-based approach, the kind I like to use to solve PS questions like this one in under a minute. Consider a few facts:
1) There are 6 married couples, so there are 12 people in all.
2) Four
people are selected. We are not even told whether these people are males or females (or, for that matter, the nature of these married relationships).
For the sake of simplicity, think of a traditional married couple that is composed of a male and a female. This means the overall composition of the group would be MMMMMMFFFFFF. Just looking at the potential to select
unmarried couples, it seems much more likely than to land upon a correct husband-wife combination. (I mean, there are numerous ways—15, for the fact-checker out there, from 6C4—in which just four
males could be selected from the six, and this combination would be mirrored for the females.) Even assuming we choose 2 males and 2 females,
a best-case scenario, one that is not even most probable (4M/0F, 3M/1F, 2M/2F, 1M/3F, 0M/4F), there would still be a less-than-likely chance that either of these two would happen to be the partner of the other two. If we consider Female 1 in such a case, for instance, there would be a 1 in 3 probability that her mate would be one of the two men; for Female 2, if Male 1 was out of the way, the probability would be lower. All things considered, it does not seem too likely that a married couple will be selected. If you think of the question itself, it is asking about the probability that
none of them would be married to each other, so we know that, as unlikely as it may be for 2 couples out of 2 to be selected, there may be other ways of generating
at least one couple. This would increase the likelihood of selecting a married pair, but would it do so on the order of 2/3—the remaining portion of the fraction in choice (C)—or greater? In other words, would there be a 2/3 probability
or greater of selecting a married pair? I think not. I love crunching numbers, but I do not need to run through any difficult calculations to see that choices (A), (B), or (C) do not make sense.
Eliminate them.
Considering the two remaining answer choices, (E) is the outlier. Applying the same reasoning as before, we would be saying that there would be just a 1/12 probability of selecting
at least one married couple. But consider our scenarios from before:
4M/0F - No couples
3M/1F - That 1 Female has a 1/2 probability of being the wife of one of those 3 males
2M/2F - Now there are several ways to pick out a married couple
1M/3F - That 1 Male has a 1/2 probability of being the husband of one of those 3 females
0M/4F - No couples
Logic would point to an answer that is not as extreme as 11/12.
The answer must be (D). This one took no pen and board, just a little number sense and a careful reading of the question, as well as a consideration of the implications of the answer choices. If I can do it, so can you. There is nothing wrong with knowing how to use the test against itself. Nothing beats a sure answer, and one that comes in less than a minute can be a boon to your test-taking success.
Good luck with your studies.
- Andrew