Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Two keys are to be put into already exisiting 5 keys in a [#permalink]
26 Apr 2006, 06:19

Two keys are to be put into already exisiting 5 keys in a key chain. What is the probability that these two keys will be adjacent on the chain? Assume key chain to be circular.

S1 - S5 are the keys that are already on the chain.

A is the first key we put into the chain. The position doesn't matter.

After that, there are 6 slots where the second key can be put into. From these, 2 are adjacent to A.

So the probability should be 2/6 or 1/3.

I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.

I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.

The picture will ALWAYS looks like this after the first key of us has
been put into the chain? The numbers next to the other 5 keys are only
a help to count a bit quicker. And besides, what's the difference it the
other 5 keys had different arrangements?

I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.

The picture will ALWAYS looks like this after the first key of us has been put into the chain? The numbers next to the other 5 keys are only a help to count a bit quicker. And besides, what's the difference it the other 5 keys had different arrangements?

I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different.

Then Key1 at S1, Key2 at S2, Key3 at S3, Key4 at S4 and Key5 at S5 is different from Key1 at S2, Key2 at S3, Key3 at S1, Key4 at S5 and Key5 at S4... and so on...

I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different.

But these keys aren't important for the solution of the question. What
matters is that there are only 6 slots after our first key has been
inserted, and of these, 2 are adjacent to the first key. So how could the
probability be 2/7?

I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different.

But these keys aren't important for the solution of the question. What matters is that there are only 6 slots after our first key has been inserted, and of these, 2 are adjacent to the first key. So how could the probability be 2/7?

Why shouldn't these 5 keys be important to the solution.

Again, I'll label the 5 keys as: Key1, Key2, Key3, Key4, Key5. I'll label the two keys to be inserted as Key A, Key B.

The arrangement: Key1, KeyA, KeyB, Key2, Key3, Key4, Key5 is different from
The arrangement: Key1, KeyA, KeyB, Key3, Key2, Key4, Key 5 is different from
The arrangement: Key1, KeyB, KeyA, Key3, Key2, Key4, Key5 is different from
The arrangement: Key1, Key2, KeyA, KeyB, Key3, Key4, Key5.... and so on...

A different arrangement can be obtained by keeping Key A and Key B are a fixed location, and have keys 1-5 in different positiosns. Similarly, Key A and Key B can move to different locations for a fixed positioning of keys 1-5.

That's my thought on this. Bascially, if Keys1-5 are fixed and cannot be shifted, then yes, I agree with your approach. But if it means I can change the arrangement of Keys1-5, then I'll stick with my approach.

Of course, I can be wrong, no one is always right.
It's a pretty interesting question, I must say.

Of course we COULD shift the other keys to different positions, but
this is a probability question. In these, we always assume that each
case is of equal probability if not otherwise stated. If not, almost
all of these probability questions wouldn't be solvable with the
information given.

If it weren't a CIRCULAR chain, then there would be 7 spots, but it IS
a circular chain, and so, there are just 6. And you won't convince me
that 2/7 is the solution unless you showed that there are 7!

Of course we COULD shift the other keys to different positions, but this is a probability question. In these, we always assume that each case is of equal probability if not otherwise stated. If not, almost all of these probability questions wouldn't be solvable with the information given.

If it weren't a CIRCULAR chain, then there would be 7 spots, but it IS a circular chain, and so, there are just 6. And you won't convince me that 2/7 is the solution unless you showed that there are 7!

S1 - S5 are the keys that are already on the chain.

A is the first key we put into the chain. The position doesn't matter.

After that, there are 6 slots where the second key can be put into. From these, 2 are adjacent to A.

So the probability should be 2/6 or 1/3.

I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.

I think 1/3 is with the fact that the 5 keys have different arragements

I think ccax's diagram explains it pretty well. But I think that's true only if there're already a bunch of keys on the key-ring and you can't take them out of the key ring and do some re-arranging.

Still, it's good discussion and I agree with the answer of 1/3.

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

Hey Everyone, I am launching a new venture focused on helping others get into the business school of their dreams. If you are planning to or have recently applied...