Find all School-related info fast with the new School-Specific MBA Forum

It is currently 18 Sep 2014, 09:52

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Probability

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
User avatar
Joined: 01 May 2004
Posts: 336
Location: USA
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
Probability [#permalink] New post 14 May 2004, 17:20
Hello!
Guys, try to solve this problem

9 people, including 3 couples are to be seated in a row of 9 chairs. What is the probability that

a. None of the Couples are together
b. Only one couple is together
c. All the couples are together
Manager
Manager
User avatar
Joined: 07 May 2004
Posts: 184
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 5 [0], given: 0

GMAT Tests User
Re: Probability [#permalink] New post 15 May 2004, 00:02
boksana wrote:
Hello!
Guys, try to solve this problem

9 people, including 3 couples are to be seated in a row of 9 chairs. What is the probability that

a. None of the Couples are together
b. Only one couple is together
c. All the couples are together


Hi, boksana!

All combinations: 9!

Combinations for which all the couples are together:

6!*2*2*2 (because there are 6! ways to arrange 6 entities - 3 couples and 3 other persons, and 2 ways of sitting a couple MW and WM).

=> p(c) = 6!*8/9! = 1/63.

Combinations for which 2 couples are together and 1 is not:

7!*2*2 - 6!*2*2*2 (the same logic - 7 entities can be seated in 7! ways, and there are 2 ways in each couple).

=> p(.) = (7!*4-6!*8)/9! = 1/18-1/63=5/126.

Combinations for which only 1 couple is:

8!*2 - [previous 2 cases together] = 8!*2 - 7!*4.

p(b) = [8!*2-7!*4]/9! = 2/9-1/18 = 1/6.

Combinations for which there are no couples:

p(a) = 1 - p(b) - p(.) - p(c) = 1 - 1/6 - 5/126 - 1/63 = 98/126 = 7/9.
Senior Manager
Senior Manager
User avatar
Joined: 01 May 2004
Posts: 336
Location: USA
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
 [#permalink] New post 15 May 2004, 10:03
Good job!
It's all correct.
Manager
Manager
User avatar
Joined: 07 May 2004
Posts: 184
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 5 [0], given: 0

GMAT Tests User
Hi! [#permalink] New post 15 May 2004, 10:26
boksana wrote:
Good job!
It's all correct.


I know... :-)

Are there any questions? Proposals? Suggestions?

Feel free to ask if you have any.
Hi!   [#permalink] 15 May 2004, 10:26
    Similar topics Author Replies Last post
Similar
Topics:
Probability Priti 8 19 Aug 2005, 11:12
Probability mirhaque 9 29 Jun 2005, 18:06
Probability peterpan 4 18 Dec 2004, 19:21
Probability oxon 5 07 Nov 2004, 12:13
Probability.... bdoll_123 0 23 Sep 2004, 07:51
Display posts from previous: Sort by

Probability

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.