|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 15 Nov 2007
Posts: 137
Followers: 1
Kudos [?]:
23
[0], given: 2
|
Given that there are 6 married couples. If we select only 4 [#permalink]
 18 Jan 2008, 01:05
Question Stats:
62% (03:42) correct
37% (02:30) wrong based on 8 sessions
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?
|
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3608
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 240
Kudos [?]:
1323
[4] , given: 347
|
4
This post received
KUDOS
p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}or p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}or p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}
_________________
NEW! GMAT ToolKit 2 (iOS) / GMAT ToolKit (Android) - The must have GMAT prep app | PrepGame
Last edited by walker on 19 Jan 2008, 02:53, edited 3 times in total.
|
|
|
|
|
|
CEO
Joined: 29 Mar 2007
Posts: 2618
Followers: 13
Kudos [?]:
145
[1] , given: 0
|
1
This post received
KUDOS
dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? Prob: 1*10/11*8/10*6/9 = 16/33
|
|
|
|
|
|
Manager
Joined: 15 Nov 2007
Posts: 137
Followers: 1
Kudos [?]:
23
[1] , given: 2
|
1
This post received
KUDOS
walker wrote: p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}
or
p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}
or
p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33} walker, explain your combinatorics method please.
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3608
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 240
Kudos [?]:
1323
[2] , given: 347
|
2
This post received
KUDOS
dominion wrote: walker wrote: p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}
walker, explain your combinatorics method please. - we choose 4 couples of 6 ones. C^2_1 - we chose one people of 2 ones for one couple. (C^2_1)^4 - we have 4 couple and for each we choose one people of 2 ones for one couple. C^{12}_4 - the total number of combinations to choose 4 people from 12 people.
_________________
NEW! GMAT ToolKit 2 (iOS) / GMAT ToolKit (Android) - The must have GMAT prep app | PrepGame
Last edited by walker on 19 Jan 2008, 02:43, edited 1 time in total.
|
|
|
|
|
|
SVP
Joined: 04 May 2006
Posts: 1946
Schools: CBS, Kellogg
Followers: 10
Kudos [?]:
170
[0], given: 1
|
walker wrote: dominion wrote: walker wrote: p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}
walker, explain your combinatorics method please. - we choose 4 pairs of 6 ones. C^2_1 - we chose one people of 2 ones for one pair. (C^2_1)^4 - we have 4 pairs and for each we choose one people of 2 ones for one pair. C^{12}_4 - the total number of combinations to choose 4 people from 12 people. No, no, please explain the other two!, exactly the other two, thanks
_________________
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3608
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 240
Kudos [?]:
1323
[3] , given: 347
|
3
This post received
KUDOS
p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\frac{12}{12} - we choose 12 of 12. \frac{10}{11} - we choose 10=12-1(prevous choice)-1(another people out of couple) of 11=12-1(prevous choice). \frac{8}{10} - we choose 8=12-2(prevous choice)-2(another people out of couple) of 10=12-2(prevous choice). \frac{6}{9} - we choose 6=12-3(prevous choice)-3(another people out of couple) of 9=12-3(prevous choice). _______________________ p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}the same logic as for C_m^n
_________________
NEW! GMAT ToolKit 2 (iOS) / GMAT ToolKit (Android) - The must have GMAT prep app | PrepGame
|
|
|
|
|
|
Director
Joined: 25 Oct 2006
Posts: 656
Followers: 7
Kudos [?]:
106
[1] , given: 6
|
1
This post received
KUDOS
Mind-Blowing... thanks walker.
_________________
If You're Not Living On The Edge, You're Taking Up Too Much Space
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
297
[2] , given: 5
|
2
This post received
KUDOS
dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? = (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4 = 12*10*8*6 / (12*11*10*9) = 16/33
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3608
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 240
Kudos [?]:
1323
[0], given: 347
|
C^6_4 - we choose 4 couples (not people) out of 6 couples (the number of all couples) the next step: we choose one person out of each couple - C^2_1
_________________
NEW! GMAT ToolKit 2 (iOS) / GMAT ToolKit (Android) - The must have GMAT prep app | PrepGame
|
|
|
|
|
|
Manager
Joined: 02 Aug 2007
Posts: 234
Schools: Life
Followers: 3
Kudos [?]:
10
[0], given: 0
|
walker wrote: C^6_4 - we choose 4 couples (not people) out of 6 couples (the number of all couples)
the next step: we choose one person out of each couple - C^2_1 Walker, Ok, its making better sense now. Are we choosing 4 couples because that means "not 4 people", ie 12-4=8?
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3608
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 240
Kudos [?]:
1323
[2] , given: 347
|
2
This post received
KUDOS
dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - C^6_4*(C^2_1)^4look at other problem: Quote: Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team? we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team. I hope it is clearer now.
_________________
NEW! GMAT ToolKit 2 (iOS) / GMAT ToolKit (Android) - The must have GMAT prep app | PrepGame
|
|
|
|
|
|
Manager
Joined: 13 Jan 2009
Posts: 179
Schools: Harvard Business School, Stanford
Followers: 3
Kudos [?]:
14
[0], given: 9
|
Agreed with 16/33. Great explanation walker!
|
|
|
|
|
|
Manager
Joined: 04 Jan 2009
Posts: 244
Followers: 1
Kudos [?]:
7
[0], given: 0
|
dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? total ways of choosing 4 out of 12:12x11x10x9 first person can be chosen in 12 ways. The next person can be chosen in 10 ways (because we don't want spouses to be in the group). The next person can be again chosen in only 8 ways (out of the 10 people left, we have to exclude 2 whose spouses we have already selected). And the last person in 6 ways. hence, probability=12x10x8x6/(12x11x10x9) = 16/33.
_________________
-----------------------
tusharvk
|
|
|
|
|
|
Director
Joined: 29 Aug 2005
Posts: 889
Followers: 6
Kudos [?]:
90
[0], given: 7
|
x2suresh wrote: dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? = (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4 = 12*10*8*6 / (12*11*10*9) = 16/33 suresh, can you please explain the logic of this calculation? Thank you!
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
297
[1] , given: 5
|
1
This post received
KUDOS
botirvoy wrote: x2suresh wrote: dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? = (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4 = 12*10*8*6 / (12*11*10*9) = 16/33 suresh, can you please explain the logic of this calculation? Thank you! 12C1 = select any person from 6 married couples (12 person) 10C1 = select second person from remaining people and exclude the first person's spouse 8C1 = select 3rd person from remain people exclue first and second perssons's spouses 6C1 = select 4th person from remaining people exclude 1st,2nd ,3rd persons's spouse Becuase order is not matter.. you need to divide by 4! Did you get it? =
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Manager
Joined: 02 Aug 2007
Posts: 234
Schools: Life
Followers: 3
Kudos [?]:
10
[0], given: 0
|
walker wrote: dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - C^6_4*(C^2_1)^4look at other problem: Quote: Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team? we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team. I hope it is clearer now. Wow, makes very clear sense now. Much obliged. Ali
|
|
|
|
|
|
Senior Manager
Joined: 06 Jul 2007
Posts: 288
Followers: 3
Kudos [?]:
25
[0], given: 0
|
xALIx wrote: walker wrote: dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - C^6_4*(C^2_1)^4look at other problem: Quote: Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team? we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team. I hope it is clearer now. Wow, makes very clear sense now. Much obliged. Ali another way of looking at this problem : total number of ways = 12C4 = 495 now let's try to find out total number of unfavourable ways. We can subtract those from 495 to get the number of favourable ways. number of ways of choosing 4 people such that there are two couples = 6C2 = 15 number of ways of choosing 4 people such that there is only one couple = number of ways of choosing one couple * number of ways of choosing two people who are not couples = 6C1 (10C2 -5) = 240. total number of unfavourable ways = 240 + 15 = 255 total number of favourable ways = 495 - 255 = 240 required probability = 240/495 = 16/33
|
|
|
|
|
|
Manager
Joined: 27 Oct 2008
Posts: 188
Followers: 1
Kudos [?]:
42
[0], given: 3
|
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?
Soln:
= (12/12) * (10/11) * (8/10) * (6/9)
= 16/33
|
|
|
|
|
|
Intern
Joined: 17 Jan 2010
Posts: 31
Schools: SSE, LSE
Followers: 0
Kudos [?]:
3
[0], given: 3
|
Thanks to the Math book on this side this was an easy one...
Thanks
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
