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Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
GRE Math Challenge #13 - Six people are asked to sit down in
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19 Sep 2014, 02:39
19 Sep 2014, 02:39
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Question Stats:
10% (01:26) correct
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Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.
_________________
Show: :: Answer
P = 6/7
Number of arrangements where at least one wife sits next to her husband:
Number of arrangements where at least one wife sits next to her husband:
_________________
Sandy
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Re: GRE Math Challenge #13 - Six people are asked to sit down in
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04 Jul 2017, 08:53
04 Jul 2017, 08:53
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sandy wrote:
Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.
Show: :: Answer
P = 6/7
Number of arrangements where at least one wife sits next to her husband:
Number of arrangements where at least one wife sits next to her husband:
Using complement rule
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 matching couple) = 1 - P(zero matching couple)
P(zero matching couple) =6/8 x 4/7 x 2/6
= 1/7
So, P(at least 1 couple) = 1 - P(no couple)
= 1 - 1/7
= 6/7
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General Discussion
Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Re: GRE Math Challenge #13
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26 Sep 2014, 01:38
26 Sep 2014, 01:38
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The question states at least one partner sits next to other in the circular arrangement. Now Initially we had to arrange 8 objects in a circular formation. 6 persons(distinguishable) and two vacant spaces (non distinguishable). So number of arrangements would be 8!/8*2! = 2520.
Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.
Divide to obtain probability: 6/7.
_________________
Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.
Divide to obtain probability: 6/7.
_________________
Sandy
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Re: GRE Math Challenge #13
[#permalink]
02 Jul 2017, 10:54
02 Jul 2017, 10:54
1
sandy wrote:
The question states at least one partner sits next to other in the circular arrangement. Now Initially we had to arrange 8 objects in a circular formation. 6 persons(distinguishable) and two vacant spaces (non distinguishable). So number of arrangements would be 8!/8*2! = 2520.
Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.
Divide to obtain probability: 6/7.
Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.
Divide to obtain probability: 6/7.
won't the two people who form the couple arrange among themselves in 2 ways?
Re: GRE Math Challenge #13 - Six people are asked to sit down in
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03 Jul 2017, 01:26
03 Jul 2017, 01:26
Expert Reply
Waiting the experts
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Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
04 Aug 2020, 06:53
04 Aug 2020, 06:53
I am unable to understand this 
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Ever Tried? Ever Failed? No Matter. Try Again. Fail Again. Fail Better!!
Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
10 Nov 2020, 08:23
10 Nov 2020, 08:23
5
I believe this could be solved very simply following this path:
probability that one wife will sit next to her husband is 1/7. Since the arrangement is circular, the husband can be on either side of his wife, so probability increases by 2: 1/7x2=2/7
Since there are three couples, probability increases by 3: 2/7x3=6/7
Hence, the answer is 6/7
probability that one wife will sit next to her husband is 1/7. Since the arrangement is circular, the husband can be on either side of his wife, so probability increases by 2: 1/7x2=2/7
Since there are three couples, probability increases by 3: 2/7x3=6/7
Hence, the answer is 6/7
Re: GRE Math Challenge #13
[#permalink]
20 Nov 2020, 06:55
20 Nov 2020, 06:55
sandy wrote:
Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.
You forgot to multiply by 3.
Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
10 Dec 2020, 12:46
10 Dec 2020, 12:46
this type of problem come in gre exam??
Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
10 Dec 2020, 13:09
10 Dec 2020, 13:09
Expert Reply
void wrote:
this type of problem come in gre exam??
Could be. It is a bit off but legit
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Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
20 Dec 2020, 15:20
20 Dec 2020, 15:20
3
pranab223 wrote:
sandy wrote:
Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.
Show: :: Answer
P = 6/7
Number of arrangements where at least one wife sits next to her husband:
Number of arrangements where at least one wife sits next to her husband:
Using complement rule
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 matching couple) = 1 - P(zero matching couple)
P(zero matching couple) =6/8 x 4/7 x 2/6
= 1/7
So, P(at least 1 couple) = 1 - P(no couple)
= 1 - 1/7
= 6/7
Hi, Thank you for explanation, I am trying to understand how your equation meets which use cases.
It does not change conclusion, but, should' P(zero matching couple) be, = 5/7 x 3/5 x 1/3 = 1/7 ?
#1st couple, (F1,F2)
F1 seat can be anything from 8 empty seats.
Remaining 7 (8-1) seats. (?/7)
F2 cannot be next to F1, so need to be picked from 5 seats (7-2)
5/7
#2nd couple, (S1,S2)
S1 seat can be anything from remaining 6 seats. (6 = (8-2) as 2 seats are taken by F1,F2)
Remaining 5(6-1) seats. (?/5)
S2 cannot be next to S1, so need to be picked from 3 seats (5-2)
3/5
#3rd couple (T1, T2)
T1 seat can be anything from remaining 4 seats. (4 = (8-4) as 4 seats are taken by F1,F2, S2,S2)
Remaining 3 (4-1) seats (?/3)
T2 cannot be next to T1,so need to be picked from 1 seats (3-2)
1/3
5/7 * 3/5 * 1/3 = 1/7
I am trying to understand whether above equation means same concept which you explained.
Thank you in advance.
Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
06 Jul 2023, 05:10
06 Jul 2023, 05:10
pranab223 wrote:
sandy wrote:
Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.
Show: :: Answer
P = 6/7
Number of arrangements where at least one wife sits next to her husband:
Number of arrangements where at least one wife sits next to her husband:
Using complement rule
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 matching couple) = 1 - P(zero matching couple)
P(zero matching couple) =6/8 x 4/7 x 2/6
= 1/7
So, P(at least 1 couple) = 1 - P(no couple)
= 1 - 1/7
= 6/7
Please how did you get the 6/8 * 4/7 * 2/6?
Can you please explain?
Re: GRE Math Challenge #13 - Six people are asked to sit down in
[#permalink]
06 Jul 2023, 06:10
06 Jul 2023, 06:10
Expert Reply
8 sits and 6 people and 3 couples
Pick one and you have 6 over 8 = 6/8
Pick the second and you have 2 couples over one less sit 4/7
Pick third and you have with the same logic 2/6
Multiply together and you have 1/7 which is the event that will NOT happen.
using 1-1/7= and you will have all the events that will happen 6/7
_________________
Pick one and you have 6 over 8 = 6/8
Pick the second and you have 2 couples over one less sit 4/7
Pick third and you have with the same logic 2/6
Multiply together and you have 1/7 which is the event that will NOT happen.
using 1-1/7= and you will have all the events that will happen 6/7
_________________
gmatclubot
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