Round table seating probability with 8 people [duplicate]
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This question is an exact duplicate of:
Probability. About round table
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Mr. A, Mr. B and $6$ more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe $(Ω,F,P)$
I think n=(8−7)!=5040 than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham m=(7−1)!∗2=1440 Than P=m/n=1440/5040=2/7 Is ir right?
And how I should describe (Ω,F,P)?
probability probability-theory
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
asked Nov 13 at 22:17
Atstovas
284
marked as duplicate by Donald Splutterwit, Lee David Chung Lin, Rebellos, user10354138, Chinnapparaj R Nov 14 at 4:35
This question was marked as an exact duplicate of an existing question.
add a comment |
up vote
0
down vote
favorite
This question is an exact duplicate of:
Probability. About round table
1 answer
Mr. A, Mr. B and $6$ more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe $(Ω,F,P)$
I think n=(8−7)!=5040 than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham m=(7−1)!∗2=1440 Than P=m/n=1440/5040=2/7 Is ir right?
And how I should describe (Ω,F,P)?
probability probability-theory
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
asked Nov 13 at 22:17
Atstovas
284
marked as duplicate by Donald Splutterwit, Lee David Chung Lin, Rebellos, user10354138, Chinnapparaj R Nov 14 at 4:35
This question was marked as an exact duplicate of an existing question.
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is an exact duplicate of:
Probability. About round table
1 answer
Mr. A, Mr. B and $6$ more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe $(Ω,F,P)$
I think n=(8−7)!=5040 than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham m=(7−1)!∗2=1440 Than P=m/n=1440/5040=2/7 Is ir right?
And how I should describe (Ω,F,P)?
probability probability-theory
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
asked Nov 13 at 22:17
Atstovas
284
This question is an exact duplicate of:
Probability. About round table
1 answer
Mr. A, Mr. B and $6$ more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe $(Ω,F,P)$
I think n=(8−7)!=5040 than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham m=(7−1)!∗2=1440 Than P=m/n=1440/5040=2/7 Is ir right?
And how I should describe (Ω,F,P)?
This question is an exact duplicate of:
Probability. About round table
1 answer
probability probability-theory
probability probability-theory
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
asked Nov 13 at 22:17
Atstovas
284
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
asked Nov 13 at 22:17
Atstovas
284
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
edited Nov 14 at 4:35
Chinnapparaj R
4,4431725
4,4431725
asked Nov 13 at 22:17
Atstovas
284
asked Nov 13 at 22:17
Atstovas
284
asked Nov 13 at 22:17
Atstovas
284
284
marked as duplicate by Donald Splutterwit, Lee David Chung Lin, Rebellos, user10354138, Chinnapparaj R Nov 14 at 4:35
This question was marked as an exact duplicate of an existing question.
marked as duplicate by Donald Splutterwit, Lee David Chung Lin, Rebellos, user10354138, Chinnapparaj R Nov 14 at 4:35
This question was marked as an exact duplicate of an existing question.
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04
add a comment |
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04
add a comment |
1 Answer
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up vote
0
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You can get 2/7 directly. Seat A anywhere, then there are 2 out of 7 places for B to be next to A.
answered Nov 13 at 22:40
herb steinberg
2,1582310
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can get 2/7 directly. Seat A anywhere, then there are 2 out of 7 places for B to be next to A.
answered Nov 13 at 22:40
herb steinberg
2,1582310
add a comment |
up vote
0
down vote
You can get 2/7 directly. Seat A anywhere, then there are 2 out of 7 places for B to be next to A.
answered Nov 13 at 22:40
herb steinberg
2,1582310
add a comment |
up vote
0
down vote
up vote
0
down vote
You can get 2/7 directly. Seat A anywhere, then there are 2 out of 7 places for B to be next to A.
answered Nov 13 at 22:40
herb steinberg
2,1582310
You can get 2/7 directly. Seat A anywhere, then there are 2 out of 7 places for B to be next to A.
answered Nov 13 at 22:40
herb steinberg
2,1582310
answered Nov 13 at 22:40
herb steinberg
2,1582310
answered Nov 13 at 22:40
herb steinberg
2,1582310
answered Nov 13 at 22:40
herb steinberg
2,1582310
2,1582310
add a comment |
add a comment |
Something I really dislike about these kind of questions, just because the only measure you have is "seating positions are uniformly sampled" doesn't mean thats the "truth". Now Mr A and Mr B don't exist, but if they did their seating probabilities wouldn't be uniform.
– s.harp
Nov 13 at 23:04