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 500,000 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:

There are 7 keys in a key ring. If two more keys are to be [#permalink]

Show Tags

01 Aug 2008, 02:46

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

the two new keys are 1 and 2

there are 7 places for the key no 1 to go and it can go in any place. probability = 7/7

for second key, there are 8 places, and it can go only in two places, left ot right of key no 1, probability = 2/8 we can select first key in two ways, probability = 1/2 final probablity = 7/7 * 2/8 * 1/2 = 1/8 answer ...

EDIT : highlighted step is not required. the answer should be : 7/7 * 2/8 = 1/4

Last edited by durgesh79 on 01 Aug 2008, 06:20, edited 1 time in total.

Once you add the first key, they are now 8 => 8 possible places for the second key to go.

Why do you insert one key before the other ? I do not see it written anywhere.

What I say is that the two possibilities (left and right of the first inserted key) could just as well be counted as one.

There are 7 places where to place the 2 keys, it is not by dividing a place by 2 (with your first key) that you should be able to change the probability of the second key.

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

number of ways 9 keys can be arranged in a ring = (9-1)! = 8! if we consider two keys as a unit, number of ways 8 keys can be arranged = (8-1)! = 7! these two keys can interchange their postions in 2 ways

@ maratikus: if you have a different approach, please tell me details

There are 7 ways of putting two keys next to each other -> put both of them in front of key1, key2, ..., key7. There are 28 ways of adding two keys. 7 places for key new1, 8 places for key new2 (order doesn't matter -> divide by 2). 7/28 = 1/4 -> D

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

the two new keys are 1 and 2

there are 7 places for the key no 1 to go and it can go in any place. probability = 7/7

for second key, there are 8 places, and it can go only in two places, left ot right of key no 1, probability = 2/8 we can select first key in two ways, probability = 1/2 final probablity = 7/7 * 2/8 * 1/2 = 1/8 answer ...

EDIT : highlighted step is not required. the answer should be : 7/7 * 2/8 = 1/4

Hi durgesh,

Could you explain why the answer below is wrong? 1 x 2/8

number of ways 9 keys can be arranged in a ring = (9-1)! = 8! if we consider two keys as a unit, number of ways 8 keys can be arranged = (8-1)! = 7! these two keys can interchange their postions in 2 ways

required probability = 2*7! / 8! = 1/4

At first, I had the same approach which I often use, but I still thought 9 keys arranged in a line >>> the number of ways can be arranged in a line = 9! the number of ways 2 keys can be arranged adjacent = 2!*8! and the answer = 2!*8!/9! = 2/9 So I was confused, I only guested 1/8 Hi, thank you, durgesh79, here is a ring, not a line

@ maratikus: if you have a different approach, please tell me details

There are 7 ways of putting two keys next to each other -> put both of them in front of key1, key2, ..., key7. There are 28 ways of adding two keys. 7 places for key new1, 8 places for key new2 (order doesn't matter -> divide by 2). 7/28 = 1/4 -> D

Thanks maratikus, I try to understand, but the explanation is not clear. Please again

arjtryarjtry !! I'm curious to know the OA... Does that reaction of yours mean OA is 1/8 Could you please confirm the OA, becuase now we are getting 1/4

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

IMO D)

Probability of two keys to be adjacent = Probability when 2nd key is added it is adjacent to 1st key added = 2/Number of spaces = 2/8 = 1/4

I think the answer makes us assume that you place the keys onto the ring 1 at a time. We start with 7, and add 1. This one doesn't matter where it goes so it is not factored into the probability.

The second key does matter because we need to figure out where it can go and what is the chance it will be next to (means on either side) of the key we just put on the ring. We now have 8 keys on the ring.

We must place the 2nd key into a space between the keys, so we actually have 9 spaces. Look below: (let the O represent an existing key on the ring).

|O|O|O|O|O|O|O|O|

However, this is not entirely correct because the questions states the keys are on a ring, not laid out on a flat surface it appears to be above with the |O|. So the first and last | is actually the same because the ring is circular. This gives us 8 spots the key can go. If the new key (first one we placed) is represented by O, then there are 2 places that will satisfy the question of the keys being placed together. That is 2 places out of 8. so 2/8 probability reduces to 1/4 (D).

arjtryarjtry wrote:

. There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent? (A) 1/8 (B) 1/6 (C) 1/5 (D) 1/4 (E) 1/3

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.