M06-10

I was thinking to solve it in the following way that is not correct. Please do let me know what is wrong in this approach.

Prob: No. of ways to select 4 persons who are not married to each other / No. of ways to select 4 people out of 12.

No. of ways to select 4 persons who are not married to each other: 12 * 10 * 8 * 6 (For the first person to be selected there are 12 options, once he is selected then for the second person there are 10 options and so on)

No. of ways to select 4 people out of 12 = (12 * 11 * 10 * 9) / (4 * 3 * 2) = 11 * 5 * 9

Now, when I divide (12 * 10 * 8 * 6) and (11 * 5 * 9) I don't get the answer. Could you please let me know what is wrong in my approach.

Hi,
The top 12*10*8*6 is not selection it is arrangement.
Bottom should also be arrangement then that is 12!/8!

Else make top selection 12*10*8*6/4!
And bottom selection 12!/4!*8!

Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection.
Hope I am clear

Hi,
The top 12*10*8*6 is not selection it is arrangement.
Bottom should also be arrangement then that is 12!/8!

Else make top selection 12*10*8*6/4!
And bottom selection 12!/4!*8!

Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection.
Hope I am clear

For the less mathematically inclined individual, logic, one of my favorite tools on the Quant section, can help you arrive at the correct solution in a time-efficient manner. Consider that if 4 people are selected from 6 married (presumably monogamous) couples, or 12 people, then at best, you would pick out 2 separate couples--just one-third of the total couples. It is far more likely that fewer couples would be selected. Eliminate choices (A), (B), and (C). In the 50/50 scenario that remains, we have a little under half a chance versus a virtual guarantee that a couple will not be selected. But choice (E) is a little too extreme. If you consider it in reverse, it is saying that there is just a single chance in 12 that a married couple would be selected, and we know this cannot be true. If any given person were selected, then there would be a 1/11 probability of selecting that person's mate, NOT a 1/12 probability. Thus, choice (E) can be burned off, and one answer choice, the correct one, remains. Do not overlook the possibility of using the test against itself.

Happy studies.

- Andrew

chetan2u ScottTargetTestPrep JeffTargetTestPrep

Can you help explain this question? Thank you!!

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

Hi,

The method is pretty straight if you realise what is happening.
There are 6 couples, so 12 in total.
Pick the first one - It can be any of these 12 person, say A in one such case.
Pick the second one - You have 11 to pick up from. The restriction is that you cannot pick up a couple, so you can not pick up the better half of A. That is remove the couple from list. You have 5 couple or 10 person left.
Pick the third one - similarly remove the two couples that have been picked in the first two picks. You are left with 4 couples or 8 to pick from.
Pick the fourth one - you are left with 3 couples or 6 person to pick from.
Total ways keeping restriction in mind = 12*10*8*6

Over all ways - 12*11*10*9

Probability = (12*10*8*6)/(12*11*10*9)=(8*6)/(11*9)=(8*2)/(11*3)=16/33

Solution:

I’ll provide two approaches for this question.

In the first approach, notice that there are 6 x 2 = 12 total people and there are 12C4 = 12!/(4!*8!) = (12 x 11 x 10 x 9)/(4 x 3 x 2) = 11 x 5 x 9 ways to pick 4 people from a total of 12 people, without any restrictions. Let’s determine the number of ways to pick 4 people with the restriction that no two of the four people are spouses. We can choose 4 couples out of 6 in 6C4 = 6!/(4!*2!) = (6 x 5)/2 = 15 ways. For each couple, we can choose the wife or the husband, which gives us 2 ways per couple; thus the number of ways we can choose 4 people where no two of them are married is 15 x 2 x 2 x 2 x 2. Thus, the required probability is (15 x 2 x 2 x 2 x 2)/(11 x 5 x 9) = (2 x 2 x 2 x 2)/(11 x 3) = 16/33.

In the second approach, let the first person be chosen. Of the remaining 11 people, 10 are not the spouse of the first person, so the probability that the second chosen person is not married to the first is 10/11. Of the remaining 10 people, 8 are not the spouse of the first or second person, so the probability that the third person is not married to the first or second is 8/10. Finally, of the remaining 9 people, 6 are not married to the first, second or third person; thus the probability that the fourth person is not married to the first, second and third is 6/9. The probability that the selection does not contain any married couples is (10/11)x(8/10)x(6/9) = (8 x 2)/(11 x 3) = 16/33.

Answer: D

I think this is a high-quality question.
I wasn't able to answer it correctly during the test - even burned through 4 min on it. But, the next day while reviewing the test, I was able to answer it easily enough. Strange, no?

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and I agree with explanation.

Bunuel KarishmaB

Would it be possible to solve it using the following approach? However, I am arriving at a wrong answer even though logically it makes sense:

Probability (None) = 1 - Probability (All)

P(all) = 6C2 (Selection Two Couples) * 2C2 (Select both members of first couple) * 2C2 (Select both members of second couple) / 12C4
P(all) = 2 /33

P (none) = 1 - 2/33 = 31/33

Can you please help me understand on which step am I faltering?

We are not allowed the case where 2 couples are selected so you subtracted that out of 1.

How about the case where exactly 1 couple is selected along with 2 other people? We need to subtract that out too.

So we calculate the probability of selecting at least 1 couple as

\(\frac{(6C1*10C2 - 6C2) }{ 12C4} = \frac{17}{33} \)

I like the solution - it’s helpful.

I like the solution - it’s helpful.