Last visit was: 27 Apr 2026, 03:00 It is currently 27 Apr 2026, 03:00
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
techieGuy
User avatar
Joined: 26 Apr 2006
Last visit: 22 Aug 2006
Posts: 35
Own Kudos:
3
Posts: 35
Kudos: 3
GMATPrep : Probability 13 Aug 2006, 19:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PLease explain



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
died4me
User avatar
Joined: 03 Aug 2006
Last visit: 31 Jul 2014
Posts: 415
Own Kudos:
9
Posts: 415
Kudos: 9
GMATPrep : Probability 13 Aug 2006, 20:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think the answer is 1/3

The total possibilities are 24
4x3x2

If only one address matches w/ one envelop, the other three match-ups should be incorrect.

Let's say we have A,B,C,and D envelops and A1,B2,C3, and D4 addresses.
If A-A1, then we have two options of matching the other three envelops and 3 addresses: 1) B-C3, C-D4, D-B2 or 2) B-D4, C-B2, D-C4

Thus, 2 options per each of four exact matches generates 8 possibilities.

8/24 = 1/3.

Please correct me if I am wrong. Thanks
User avatar
bz9
User avatar
Joined: 12 Feb 2006
Last visit: 17 Jan 2010
Posts: 70
Own Kudos:
267
Posts: 70
Kudos: 267
GMATPrep : Probability 13 Aug 2006, 21:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I did it this way, but I'm probably wrong.

3/4 x 2/3 x 1/2 = 1/4

Man I am terrible at prob questions,
User avatar
prashrash
User avatar
Joined: 09 Jul 2006
Last visit: 25 Nov 2006
Posts: 53
Own Kudos:
2
Location: India
Posts: 53
Kudos: 2
GMATPrep : Probability 14 Aug 2006, 00:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
died4me
I think the answer is 1/3

The total possibilities are 24
4x3x2

If only one address matches w/ one envelop, the other three match-ups should be incorrect.

Let's say we have A,B,C,and D envelops and A1,B2,C3, and D4 addresses.
If A-A1, then we have two options of matching the other three envelops and 3 addresses: 1) B-C3, C-D4, D-B2 or 2) B-D4, C-B2, D-C4

Thus, 2 options per each of four exact matches generates 8 possibilities.

8/24 = 1/3.

Please correct me if I am wrong. Thanks
Show more


Could you please explain how you arrived at the total number of possibilities as 24 ?


I have worked it out this way:
let us assume that the first letter goes into the correct envelope.
The probability of this happening would be 1/4.

For the next letter, out of 3 possible envelopes, it can go into 2 wrong envelopes.
So probablility of letter number 2 going into WRONG envelope = 2/3.

Similarly for letter number 3, probability of going into wrong envelope is = 1/2 (since we have only 2 unfilled envelopes remaining).

Now, if the first letter has gone in the correct envelope, and the 2nd and 3rd letters have gone in the wrong envelopes, then the 4th letter will definitely go in the wrong envelope.
Therefore probability of envelope being wrong for 4th letter = 1/1.

Since all these events are dependent, the total probability would be
1/4*2/3*1/2*1/1 = 1/12.

Please let me know what is wrong in my approach to the problem.
Am facing a lot of difficulties with probability in my preparation. So any help would be highly appreciated.
Thank you.
User avatar
jaynayak
User avatar
Joined: 30 Mar 2006
Last visit: 07 Jul 2008
Posts: 893
Own Kudos:
647
Posts: 893
Kudos: 647
GMATPrep : Probability 14 Aug 2006, 00:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1/3

Number of ways four letters can be arranged = 4! = 24

Lets say A B C and D are the four letters. And they have to go into
[1] [2] [3] [4] envelopes.

When only letter A goes to the correct envelope. The other letters can be arrange in 2 ways. CDB and DBC

Similarily when only one of the four letters go into the right envelope, other letters can be arranged in 8 ways.

Hence P=8/24 = 1/3



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!