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Probability - Couples

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Probability - Couples

by Tryingmybest » Wed Mar 25, 2009 5:48 am
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -[spoiler]D bit I get 2/15[/spoiler]
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by cm47323 » Thu Mar 26, 2009 10:35 pm
Any source of this? A bit long winded to solve. There are 5 people: couple A1,A2, couple B1,B2, single C. C is the pivot person, for lack of a better term. We need to generate the separate possibilities based on Cs position.

- C is 1st. 4 choices for Seat 2, 2 choices for Seat 3 (one of the members of the other couple) and the rest of the seats are determined. (4*2)=8
- C is 5th. 8 choices also. Solved same as above
- C is 2nd. 4 choices for Seat 1. 2 choices for Seat 3, since it cannot Seat 1s partner (forces 2nd couple to sit in 4 and 5. (4*2)=8
- C is 4th. 8 choices.
- C is 3rd. Seat 1 has 4 choices. Seat 2 has 2 choices (member of other couple). Seat 4 has 2 choices and Seat 5 is determined. (4*2*2)=16
So (8+8+8+8+16)/5!= 48/120

I tried solving this other ways to no avail. It is easy to figure out #ways both couples are next to each other, but to figure #ways exactly 1 couple is next to each other is not apparent to me.
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by Vemuri » Fri Mar 27, 2009 4:48 am
I am sure there should be a easier way to do it? Is cramya around?
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by mals24 » Fri Mar 27, 2009 5:42 am
Well there is another way of doing it.

1 - Either = Neither

or for this question,

Prob. that neither couples sit together = 1 - prob. that either of the couples sit together.

Let AB - first couple
CD - second couple
E - single fellow.

P(A or B) = P(A)+P(B)-P(A and B)

P(A) = prob. of first couple sitting together.
P(B) = prob. of 2nd couple sitting together.
P(A and B) = prob. of both couples sitting together

Take AB as one unit. So the number of ways of arranging 4 units = 4!.
AB can be arranged in 2! ways.
Hence number of ways first couple can sit together = 2*4! = 48

Same for couple CD.
Number of ways 2nd couple can sit together = 2*4! = 48

Number of ways both couples can sit together.
Same logic as above. Consider couple AB and couple CD as 2 units.
So number of ways of arranging 3 units = 3!
AB and CD can be arranged in 2*2 ways (AB in 2 ways and CD in 2 ways)
both couples can sit together in 2*2*3! = 24 ways

Put it into the formula.
(A or B) = A + B - (A and B)
48+48-24 = 72

Total number of ways of arranging 5 people = 5! = 120

So prob. that either couples sit together = 72/120 = 3/5
Prob. that neither couples sit together = 1-3/5 = 2/5

This method looks long but once you understand the logic it hardly takes a minute to solve.
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Re: Probability - Couples

by Brent@GMATPrepNow » Fri Mar 27, 2009 6:11 am
Tryingmybest wrote:Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -[spoiler]D bit I get 2/15[/spoiler]
I removed my woefully erroneous post.

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Sun May 19, 2013 6:10 am, edited 1 time in total.
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by Vemuri » Fri Mar 27, 2009 6:32 am
cm47323 wrote:It is easy to figure out #ways both couples are next to each other, but to figure #ways exactly 1 couple is next to each other is not apparent to me.
How should we solve if the question is saying that the couple should be seated together?

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Re: Probability - Couples

by vittalgmat » Sat Mar 28, 2009 5:25 pm
Brent Hanneson wrote:
Tryingmybest wrote:Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -[spoiler]D bit I get 2/15[/spoiler]
# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5

Awesome explanation Brent.. many thanks!!!
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Re: Probability - Couples

by Vemuri » Sat Mar 28, 2009 9:14 pm
Brent Hanneson wrote:
Tryingmybest wrote:Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -[spoiler]D bit I get 2/15[/spoiler]
# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5
What are we finding here? The probability of couple sitting together or not sitting together? Brent can you explain your approach? I am not good in probability & so your explanation would be really helpful.
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still having a tough time with this one

by chix0r » Sat Mar 28, 2009 9:41 pm
Can someone please explain this to me a little more? I keep getting 1/5 as the answer but everyone else is getting 2/5.

I keep thinking that if there are 48 ways to seat couple A together, and 48 ways to seat couple B together, 48+48 is 96. 120-96 is 24, 24/120 = 1/5.

Can someone explain in words/logic, not in formulas, why this thinking is wrong? Thanks.
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by gmat740 » Sat Mar 28, 2009 10:11 pm
Tryingmybest wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -D bit I get 2/15


# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5


@ Brent:

What Have you calculated!!!

take a look at the bold face


The probability of couple sitting together is 24

So the number of ways of not sitting together will be
( total ways- ways in which they sit together)

60-24 = 36


So probablity is 36/60

= 3/5

Please check the OA
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by Brent@GMATPrepNow » Sun Mar 29, 2009 2:17 pm
gmat740 wrote:
Tryingmybest wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -D bit I get 2/15


# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5


@ Brent:

What Have you calculated!!!

take a look at the bold face


The probability of couple sitting together is 24

So the number of ways of not sitting together will be
( total ways- ways in which they sit together)

60-24 = 36


So probablity is 36/60

= 3/5

Please check the OA
You're 100% correct.
That's the second question this weekend that I didn't read it through.
I calculated the probability that the couple IS together.
So, we need to subtract that from 1 to get 3/5
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by Vemuri » Sun Mar 29, 2009 11:25 pm
Brent Hanneson wrote:You're 100% correct.
That's the second question this weekend that I didn't read it through.
I calculated the probability that the couple IS together.
So, we need to subtract that from 1 to get 3/5
But, the answer choice 3/5 is not in the options.

Can someone elaborate the approach a bit further? I am a dummy in probability & want to understand how to solve this problem.
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Re: Probability - Couples

by Vemuri » Mon Mar 30, 2009 7:41 am
I am going to pull this question back up because no one has been able to solve this problem convincingly. Let me give a shot at it myself with my novice knowledge on probability.

To find the probability that neither of the couples sits together in adjacent chairs, we have to find the probability that they sit together in adjacent chairs & then subtract from the whole to get the answer.

Lets assume the couples are 'AB' & 'CD' & the single person is E. The number of ways the couples can be seated adjacent & the single person can be seated = 3! 2! 2! = 24 ways

Total number of ways possible = 5! = 120 ways.

The probability that the couples sit together = 24/120 = 1/5

So, the probability that the couples do not sit together = 4/5

This is not an option in the answer choices. Am I doing something wrong or are the answer choices not correct?
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by gmat740 » Mon Mar 30, 2009 3:52 pm
Hello Vemuri.
Sorry for late reply, I was quiet busy for the last couple of days
Total number of ways possible = 5! = 120 ways.

The probability that the couples sit together = 24/120 = 1/5

So, the probability that the couples do not sit together = 4/5

Well What have you calculated is the arrangement of chairs and not the people!!


Let [] ==> denotes each chair

Let the couple be denoted by C1 and C2 and the other single person By S

Please draw 6 chairs in some scratch paper for better understanding

[] [] [] [] [] []

Lets first sit C1
[C1] [] [] [] [] []

Now C2 can take only the adjacent seat

[C1] [C2] [] [] [] []

now S can seat in 4 different ways(4 empty chairs!)

Although, C1 and C2 can also be interchanged as C1 C2 or C2 C1 which I will deal latter in this question

Similarly,
for another Seating arrangement like this

[] [C1] [C2] [] [] []

Now again S can be seated in 4 different ways,one on left of C1 and 3 on right of C2

[] [] [C1] [C2] [] []
Again 4 types of seating arrangements for S


Now if we further advance C1 C2 to the right, the Second arrangement again occurs

[] [] [] [C1] [C2] []

So we are not going to count this arrangement.

So at Max, there can be 3 different arrangements(Bold Face) having 4 types of seating arrangement of S

thus, 3*4 = 12
Further, this seating arrangement of C1 C2 Can be interchanged as C2 and C1

12*2 = 24

And Total Number of Arrangements of 3 people in 6 chairs is

6*5*4/2! =120/2 = 60
two people as couples(same type) = 2!

So ways of couple not seating togather is 60-24 =36

So probability = 36/60 =3/5

Hope now everything is clear!!
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by Vemuri » Mon Mar 30, 2009 5:13 pm
Thank you for replying GMAT740. I have a retraced your explanation with a correction:

The question mentions only 5 chairs:

[C1] [C2] [] [] [] --> S can seat in 3 different ways.
[] [C1] [C2] [] [] --> S can seat in 3 different ways.
[] [] [C1] [C2] [] --> S can seat in 3 different ways.
[] [] [] [C1] [C2] --> S can seat in 3 different ways.

Hence, the arrangements = 4 X 3 X 2 = 24

Total possible arrangements is 5!/2! (the 2! is the arrangement of C1 & C2, i.e. C1C2 or C2C1) ==> 120/2 = 60

The probability of the couples sitting together is 24/60 = 2/5
Hence, the probability of the couples not sitting together is 1-2/5 = 3/5.

Is this the correct answer? TryingmyBest, can you please check the answer options again? What is the source of this question?
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