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There are 4 couples. You arrange these people in a line.

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There are 4 couples. You arrange these people in a line. [#permalink] New post 21 Aug 2003, 14:51
There are 4 couples. You arrange these people in a line. What is
the probability that no husband is with his wife.

While i got to the answer, it doesnt match...please help

my solution and answer to follow

thanks
praetorian
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 [#permalink] New post 21 Aug 2003, 19:41
I'm really bad at probability....anyways...is it 69/70??...by any chance?
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 [#permalink] New post 22 Aug 2003, 19:31
There are 4 couples = 8 men/women

TOTAL COMBINATIONS = 8C2

TOTAL UNFAVORABLE COMBINATIONS
Number of combinations where couples are together = 4

1 - (4)/(8C2) = 6/7

Is this right ?
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Re: Probability # 11 [#permalink] New post 22 Aug 2003, 20:38
praetorian123 wrote:
There are 4 couples. You arrange these people in a line. What is
the probability that no husband is with his wife.

While i got to the answer, it doesnt match...please help

my solution and answer to follow

thanks
praetorian


Where did you get this problem. It is quite difficult, IMO. In any case I ended up with 11/21
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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

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Kudos [?]: 670 [0], given: 781

Re: Probability # 11...Akamai..Please Comment [#permalink] New post 22 Aug 2003, 20:53
AkamaiBrah wrote:
praetorian123 wrote:
There are 4 couples. You arrange these people in a line. What is
the probability that no husband is with his wife.

While i got to the answer, it doesnt match...please help

my solution and answer to follow

thanks
praetorian


Where did you get this problem. It is quite difficult, IMO. In any case I ended up with 11/21




This was posted in another group...he told us that its from a Probability book and as we found out, its way too hard for a GMAT problem.
But i guess its better to solve these hard conceptual problems.

Well, the answer given is 12/35

I also didnt get the answer right...Please tell me what concept i got wrong...


Calculate the possibilities where couples can be together...and
then subtract the probability from 1.
Heres my attempt :
# of ways in which 1 couple are together = 4P1=4
# of ways in which 2 couples are together = 4P2=12
# of ways in which 3 couples are together = 4P3=24
# of ways in which 4 couples are together = 4P4=24

Total # of ways in which the couple can be together = 4*12*24*24

Total # of ways of arranging 8 people = 8!
4*12*24*24 / 8! = 24/35

So the prob that no husband is with his wife = 1- 24/35 = 11/35

My Answer : 11/35 ..please comment

thanks
praetorian
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Re: Probability # 11...Akamai..Please Comment [#permalink] New post 22 Aug 2003, 21:43
praetorian123 wrote:
AkamaiBrah wrote:
praetorian123 wrote:
There are 4 couples. You arrange these people in a line. What is
the probability that no husband is with his wife.

While i got to the answer, it doesnt match...please help

my solution and answer to follow

thanks
praetorian


Where did you get this problem. It is quite difficult, IMO. In any case I ended up with 11/21




This was posted in another group...he told us that its from a Probability book and as we found out, its way too hard for a GMAT problem.
But i guess its better to solve these hard conceptual problems.

Well, the answer given is 12/35

I also didnt get the answer right...Please tell me what concept i got wrong...


Calculate the possibilities where couples can be together...and
then subtract the probability from 1.
Heres my attempt :
# of ways in which 1 couple are together = 4P1=4
# of ways in which 2 couples are together = 4P2=12
# of ways in which 3 couples are together = 4P3=24
# of ways in which 4 couples are together = 4P4=24

Total # of ways in which the couple can be together = 4*12*24*24

Total # of ways of arranging 8 people = 8!
4*12*24*24 / 8! = 24/35

So the prob that no husband is with his wife = 1- 24/35 = 11/35

My Answer : 11/35 ..please comment

thanks
praetorian


Oops. I made a bad math error. However, I know how to solve this.

Your way is not correct even though you ended up close. There is no way to justify your multiplying all of the possibilities together.

The correct solution is as follows:

Let call the couples A, B, C, and D.

We want to know the probability that NO couple are together. This is the same as calculating the probability of NOT (A or B or C or D).

However, since these events are not mutually exclusive, we need to subtract out the parts that overlap. Without going into a lot of math, the probability will look like this:

P(no couples) = 1 - (4C1*p(one particular couple) - (4C2*p(two particular couples) - (4C3*p(three particular couples) - 4C4*p(4 couples))))

(repeated application of the OR rule with events that are not mutually exclusive -- look it up. Or imagine a venn diagram and how you would count the total number inside (you cannot draw a 4 section venn-diagram on paper, so imagine it for 3 sections))

= 1 - (4*(7!*2) - (6*(6!*4) - (4*(5!*8) - 4!*16))) / 8!

= 1 - (40320 - 17280 + 3840 - 384) / 40320

= 1 - 26496/40320

= 1 - 23/35

= 12/35
_________________

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

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CEO
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Joined: 15 Aug 2003
Posts: 3470
Followers: 60

Kudos [?]: 670 [0], given: 781

Re: Probability # 11...Akamai..Please Comment [#permalink] New post 22 Aug 2003, 21:58
Thanks for the solution..i tried for a while to understand it, but couldnt see where you got your numbers from..

also, you mentioned i could not multiply the individual cases together....can you please elaborate on that...i thought that the four cases were independent and so multiplication of the four was correct..

Its been a while since i did probability..kindly bear with me.

Thanks for your help
Praetorian
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Re: Probability # 11...Akamai..Please Comment [#permalink] New post 23 Aug 2003, 13:19
Quote:
P(no couples) = 1 - (4C1*p(one particular couple) - (4C2*p(two particular couples) - (4C3*p(three particular couples) - 4C4*p(4 couples))))

(repeated application of the OR rule with events that are not mutually exclusive -- look it up. Or imagine a venn diagram and how you would count the total number inside (you cannot draw a 4 section venn-diagram on paper, so imagine it for 3 sections))

= 1 - (4*(7!*2) - (6*(6!*4) - (4*(5!*8) - 4!*16))) / 8!
= 1 - (40320 - 17280 + 3840 - 384) / 40320

= 1 - 26496/40320

= 1 - 23/35

= 12/35[/quote]


Akamai,

Can you please explain how you worked out the bold part above?

Thanks,
Re: Probability # 11...Akamai..Please Comment   [#permalink] 23 Aug 2003, 13:19
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There are 4 couples. You arrange these people in a line.

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