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

It is currently 30 Aug 2014, 20:44

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

Six married couples are standing in a room. If 4 people are

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

Six married couples are standing in a room. If 4 people are [#permalink] New post 16 Aug 2003, 02:28
Six married couples are standing in a room.
If 4 people are chosen at random, find the prob. that exactly one married couple is there among the 4 people.

how dyo solve this? id say:
we have to select only one couple so:
1c6*2c10/4c12=6/11... here i would have, however, combinations with
2 couples, so i have to substract these
2 couples=2c6/4c12=1/33

so answer=6/11-1/33=17/33...

but the official answer is 16/33 (in fact if i try to solve it by 1. finding the prob that 0 couples are among the 4 people: 12*10*8*6/4!/(4c12)=16/33; 2. finding the prob that 2 couples are among the 4 people: 2c6/4c12=1/33; and 3. prob 1 couple=1-prob 0 couples - prob 2 couples=1-16/33-1/33=16/33; i get the right anwer) ...

aaargh!!!... what am i missing???... pls help. thx, j
SVP
SVP
User avatar
Joined: 03 Feb 2003
Posts: 1613
Followers: 5

Kudos [?]: 44 [0], given: 0

 [#permalink] New post 16 Aug 2003, 13:46
my approach

FAVORABLE: a four-position group _ _ _ _

1. take any pair and fill two places (6 ways)
2. take another person to fill the third place (5 pairs=10 ways)
3. take the last person (8 ways, for we cannot take a spose of the foregoing person)

TOTAL: 12C4

(6*10*8)/(9*5*11)=32/33

I also made a mistake, but where?
Manager
Manager
avatar
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

 [#permalink] New post 16 Aug 2003, 14:24
another way i tried to solve this problem was the following:
for the first person i have 12 possibilities
for the second (assuming first two people form couple), only 1
for the third, i have 10 possibilities
for the fourth, i have 8 possibilities
so number of arrangement with one couple=(12*1*10*8)/4!

this yields ((12*1*10*8)/4!)/(12c4/4!)=8/99

aaargh :horror
Manager
Manager
avatar
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

 [#permalink] New post 16 Aug 2003, 14:35
i think i know where the problem is:
((12*1)/2!)*((10*8)/2!)/12c4=16/33 (right answer)

note the difference with ((12*1*10*8)/4!)/12c4?

however, i cant figure out why is the first approach right...
GMAT Instructor
User avatar
Joined: 07 Jul 2003
Posts: 771
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 9

Kudos [?]: 26 [0], given: 0

GMAT Tests User
Re: probability problem [#permalink] New post 19 Aug 2003, 06:43
javropu wrote:
Six married couples are standing in a room.
If 4 people are chosen at random, find the prob. that exactly one married couple is there among the 4 people.

how dyo solve this? id say:
we have to select only one couple so:
1c6*2c10/4c12=6/11... here i would have, however, combinations with
2 couples, so i have to substract these
2 couples=2c6/4c12=1/33

so answer=6/11-1/33=17/33...

but the official answer is 16/33 (in fact if i try to solve it by 1. finding the prob that 0 couples are among the 4 people: 12*10*8*6/4!/(4c12)=16/33; 2. finding the prob that 2 couples are among the 4 people: 2c6/4c12=1/33; and 3. prob 1 couple=1-prob 0 couples - prob 2 couples=1-16/33-1/33=16/33; i get the right anwer) ...

aaargh!!!... what am i missing???... pls help. thx, j


There are 12C4 = 495 ways to pick 4 people.

There are 6 ways to pick one couple, and 10C2 = 45 ways to pick the other two people, 5 of which are couples.

Hence, there are 6*(45-5)=240 ways to pick one couple in 4 people.

Hence the prob is 240/495 = 16/33
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Manager
avatar
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

 [#permalink] New post 19 Aug 2003, 08:57
thx akamay... you make it look so easy... its embarrasing... but ive learned a lot, thx... i discussed this problem with another smart guy and he told me this other way:
1c6*2c5*1c2*1c2/4c12
-you pick a couple among the 6, then you pick 2 more couples among the other 5 and then you pick a member of each couple-... cheers, javi
GMAT Instructor
User avatar
Joined: 07 Jul 2003
Posts: 771
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 9

Kudos [?]: 26 [0], given: 0

GMAT Tests User
 [#permalink] New post 19 Aug 2003, 11:12
stolyar wrote:
my approach

FAVORABLE: a four-position group _ _ _ _

1. take any pair and fill two places (6 ways)
2. take another person to fill the third place (5 pairs=10 ways)
3. take the last person (8 ways, for we cannot take a spose of the foregoing person)

TOTAL: 12C4

(6*10*8)/(9*5*11)=32/33

I also made a mistake, but where?


You double count in steps 2 and 3.

consider step #1. If you counted it the same way you would say:
1a: pick any person (12 ways)
1b: pick the spouse (1 way)
but there are only 6 pairs! If you pick the husband first then the wife, it is the same combination as if you pick the wife first, then the husband.

Similarly, you pair up each non-couple twice in steps 2 and 3 so you have exactly twice as many combinantions as you need.

Do you get it?
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Manager
avatar
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

 [#permalink] New post 19 Aug 2003, 13:03
mmmmm... ive double checked my numbers
1c6*2c5*1c2*1c2/4c12=16/33
this doesnt prove anything, i admit it...
i get your point about the first pair... however, im still missing how does that relate to my approach... ive picked one pair... 6 ways... then i pick 2 pairs out of the 5 left (to make sure i dont pick two people from the same pair)... 10 ways (not 20)... then i pick one different member of each pair... 4 ways (not 6)... so total=6*10*4=240... i just dont see where im double counting... help me see it please, as i cannot see it myself... thk u very much
GMAT Instructor
User avatar
Joined: 07 Jul 2003
Posts: 771
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 9

Kudos [?]: 26 [0], given: 0

GMAT Tests User
 [#permalink] New post 19 Aug 2003, 13:41
javropu wrote:
mmmmm... ive double checked my numbers
1c6*2c5*1c2*1c2/4c12=16/33
this doesnt prove anything, i admit it...
i get your point about the first pair... however, im still missing how does that relate to my approach... ive picked one pair... 6 ways... then i pick 2 pairs out of the 5 left (to make sure i dont pick two people from the same pair)... 10 ways (not 20)... then i pick one different member of each pair... 4 ways (not 6)... so total=6*10*4=240... i just dont see where im double counting... help me see it please, as i cannot see it myself... thk u very much


You are fine. I was responding to Stolyar.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 19 Aug 2003, 13:41
    Similar topics Author Replies Last post
Similar
Topics:
Six Married couples are standing in a room. If 4 people are neelesh 9 17 Dec 2007, 20:42
6 married couple are present at a party. If 4 people are shoonya 5 07 Aug 2007, 19:23
6 married couples are present at a party. if 4 people are shoonya 11 06 Aug 2006, 15:05
There are 4 married couples in the ball room, 4 people are conocieur 25 04 May 2006, 10:10
Six Married couples are standing in a room. If 4 people are neelesh 18 08 Jan 2005, 21:15
Display posts from previous: Sort by

Six married couples are standing in a room. If 4 people are

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.