Author 
Message 
Current Student
Joined: 11 May 2008
Posts: 551

There are 7 keys in a key ring. If two more keys are to be [#permalink]
Show Tags
01 Aug 2008, 03:46
. 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 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Director
Joined: 27 May 2008
Posts: 531

Re: probability [#permalink]
Show Tags
Updated on: 01 Aug 2008, 07:20
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 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
Originally posted by durgesh79 on 01 Aug 2008, 04:08.
Last edited by durgesh79 on 01 Aug 2008, 07:20, edited 1 time in total.



Current Student
Joined: 11 May 2008
Posts: 551

Re: probability [#permalink]
Show Tags
01 Aug 2008, 04:14
durgesh!!! good explanation..



Intern
Joined: 15 Jul 2008
Posts: 43

Re: probability [#permalink]
Show Tags
01 Aug 2008, 04:21
thank durgesh79



Senior Manager
Joined: 14 Mar 2007
Posts: 274
Location: Hungary

Re: probability [#permalink]
Show Tags
01 Aug 2008, 05:48
Just one question. Why do we need to multiple by 1/2. Can you please explain??



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: probability [#permalink]
Show Tags
01 Aug 2008, 06:31
Can you explain what is wrong with this approach ? I don't get it...
There are 7 keys on a ring, so 7 places where the keys can go to be adjacent.
You can place each of the additional 2 keys in any of these places : total number of possibilities is 7*7=49
But to have them adjacent, you have to place them in the same place : 7 possibilities
==> probability of having them adjacent is 7/49 = 1/7



Senior Manager
Joined: 16 Jul 2008
Posts: 279
Schools: INSEAD Dec'10

Re: probability [#permalink]
Show Tags
01 Aug 2008, 06:39
Oski wrote: Can you explain what is wrong with this approach ? I don't get it...
There are 7 keys on a ring, so 7 places where the keys can go to be adjacent.
You can place each of the additional 2 keys in any of these places : total number of possibilities is 7*7=49
But to have them adjacent, you have to place them in the same place : 7 possibilities
==> probability of having them adjacent is 7/49 = 1/7 The number of possibilities is 7*8 = 56 Once you add the first key, they are now 8 => 8 possible places for the second key to go.
_________________
http://applicant.wordpress.com/



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: probability [#permalink]
Show Tags
01 Aug 2008, 06:43
Nerdboy wrote: The number of possibilities is 7*8 = 56
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.



Director
Joined: 01 Jan 2008
Posts: 601

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:01
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 The answer is D. I will explain if needed.



Director
Joined: 27 May 2008
Posts: 531

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:18
zoltan wrote: Just one question. Why do we need to multiple by 1/2. Can you please explain?? now when you asked, i realized that we dont have to multiply with 1/2. The answer should be 1/4 Option D... Thanks for correcting me....



Intern
Joined: 15 Jul 2008
Posts: 43

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:22
@ maratikus: if you have a different approach, please tell me details



Director
Joined: 27 May 2008
Posts: 531

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:25
another method
number of ways 9 keys can be arranged in a ring = (91)! = 8! if we consider two keys as a unit, number of ways 8 keys can be arranged = (81)! = 7! these two keys can interchange their postions in 2 ways
required probability = 2*7! / 8! = 1/4



Director
Joined: 01 Jan 2008
Posts: 601

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:37
fiesta wrote: @ 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



Senior Manager
Joined: 19 Mar 2008
Posts: 345

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:42
durgesh79 wrote: 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 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



Director
Joined: 27 May 2008
Posts: 531

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:45
judokan wrote: Hi durgesh,
Could you explain why the answer below is wrong? 1 x 2/8 i think 1/4 is the right answer.... earlier i came up with 1/8 but later realised my mistake...



Intern
Joined: 15 Jul 2008
Posts: 43

Re: probability [#permalink]
Show Tags
01 Aug 2008, 07:59
durgesh79 wrote: another method
number of ways 9 keys can be arranged in a ring = (91)! = 8! if we consider two keys as a unit, number of ways 8 keys can be arranged = (81)! = 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



Intern
Joined: 15 Jul 2008
Posts: 43

Re: probability [#permalink]
Show Tags
01 Aug 2008, 08:07
maratikus wrote: fiesta wrote: @ 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



Director
Joined: 27 May 2008
Posts: 531

Re: probability [#permalink]
Show Tags
01 Aug 2008, 08:41
arjtryarjtry wrote: durgesh!!! good explanation.. 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



Senior Manager
Joined: 06 Apr 2008
Posts: 393

Re: probability [#permalink]
Show Tags
01 Aug 2008, 08:45
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 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



SVP
Joined: 30 Apr 2008
Posts: 1841
Location: Oklahoma City
Schools: Hard Knocks

Re: probability [#permalink]
Show Tags
01 Aug 2008, 09:28
I get D, 1/4. Here is my reason why. 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). OO OOOOOO 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$$.
GMAT Club Premium Membership  big benefits and savings







Go to page
1 2
Next
[ 22 posts ]



