What is the probablity of the following occuring

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Let's say you are at a table with 5 others. Everyone is seated random around a 6 person table. you only know 1 person at this party.
beginenumerate
item What is the likelyhood you sat next to individual that you know.
item What is the likelyhood you are sat opposite to this person that you know.
item What is the likelyhood that you sat next to two strangers.



endenumerate
So if the table is 6 people and you sat anywhere then there $5$ chairs left over. Since your friend can be seated on either side of you then that leaves 3 chairs.
So would it be 2/6 or 1/3?






probability







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  • Which question is your working for? Question 3?
    – Parcly Taxel
    1 hour ago











  • ?? 2/6 is 1/3.
    – Gerry Myerson
    1 hour ago










  • all of them. the one i was working on was #1.
    – fsdff
    1 hour ago














up vote
4
down vote

favorite












Let's say you are at a table with 5 others. Everyone is seated random around a 6 person table. you only know 1 person at this party.
beginenumerate
item What is the likelyhood you sat next to individual that you know.
item What is the likelyhood you are sat opposite to this person that you know.
item What is the likelyhood that you sat next to two strangers.



endenumerate
So if the table is 6 people and you sat anywhere then there $5$ chairs left over. Since your friend can be seated on either side of you then that leaves 3 chairs.
So would it be 2/6 or 1/3?






probability







share|cite|improve this question























  • Which question is your working for? Question 3?
    – Parcly Taxel
    1 hour ago











  • ?? 2/6 is 1/3.
    – Gerry Myerson
    1 hour ago










  • all of them. the one i was working on was #1.
    – fsdff
    1 hour ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Let's say you are at a table with 5 others. Everyone is seated random around a 6 person table. you only know 1 person at this party.
beginenumerate
item What is the likelyhood you sat next to individual that you know.
item What is the likelyhood you are sat opposite to this person that you know.
item What is the likelyhood that you sat next to two strangers.



endenumerate
So if the table is 6 people and you sat anywhere then there $5$ chairs left over. Since your friend can be seated on either side of you then that leaves 3 chairs.
So would it be 2/6 or 1/3?






probability







share|cite|improve this question















Let's say you are at a table with 5 others. Everyone is seated random around a 6 person table. you only know 1 person at this party.
beginenumerate
item What is the likelyhood you sat next to individual that you know.
item What is the likelyhood you are sat opposite to this person that you know.
item What is the likelyhood that you sat next to two strangers.



endenumerate
So if the table is 6 people and you sat anywhere then there $5$ chairs left over. Since your friend can be seated on either side of you then that leaves 3 chairs.
So would it be 2/6 or 1/3?






probability




probability






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share|cite|improve this question













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edited 10 mins ago









Naveen Gupta

302




302










asked 1 hour ago









fsdff

1363




1363











  • Which question is your working for? Question 3?
    – Parcly Taxel
    1 hour ago











  • ?? 2/6 is 1/3.
    – Gerry Myerson
    1 hour ago










  • all of them. the one i was working on was #1.
    – fsdff
    1 hour ago
















  • Which question is your working for? Question 3?
    – Parcly Taxel
    1 hour ago











  • ?? 2/6 is 1/3.
    – Gerry Myerson
    1 hour ago










  • all of them. the one i was working on was #1.
    – fsdff
    1 hour ago















Which question is your working for? Question 3?
– Parcly Taxel
1 hour ago





Which question is your working for? Question 3?
– Parcly Taxel
1 hour ago













?? 2/6 is 1/3.
– Gerry Myerson
1 hour ago




?? 2/6 is 1/3.
– Gerry Myerson
1 hour ago












all of them. the one i was working on was #1.
– fsdff
1 hour ago




all of them. the one i was working on was #1.
– fsdff
1 hour ago










3 Answers
3






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The answer to #1 is not $frac26$ but $frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $frac15$ and $frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.






share|cite|improve this answer



























    up vote
    2
    down vote













    Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem. Fix yourself, there are 2 ways your known friend can sit either side. The rest of you can sit in 4! ways to a total of 48. Thus the probability for the first question is $frac48120 = frac25$. The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$frac24120 = frac15$. The third question is the two strangers can be picked in $4choose2$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2times6times6) = 72$. Thus the probability is $frac72120 = frac35$






    share|cite|improve this answer



























      up vote
      0
      down vote













      In total, there are $6! = 720$ ways for all the people to sit on the chairs.



      1. First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$frac 24 times 2 times 6720=frac25$$


      2. Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$frac 24 times 1 times 6720=frac15$$


      3. In fact this is the complement of 1. so the probability would be $$1- frac25=frac35$$


      (Sorry, I was late and English is my second language)






      share|cite|improve this answer






















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote













        The answer to #1 is not $frac26$ but $frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $frac15$ and $frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.






        share|cite|improve this answer
























          up vote
          2
          down vote













          The answer to #1 is not $frac26$ but $frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $frac15$ and $frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.






          share|cite|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            The answer to #1 is not $frac26$ but $frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $frac15$ and $frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.






            share|cite|improve this answer












            The answer to #1 is not $frac26$ but $frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $frac15$ and $frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Parcly Taxel

            36.1k136993




            36.1k136993




















                up vote
                2
                down vote













                Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem. Fix yourself, there are 2 ways your known friend can sit either side. The rest of you can sit in 4! ways to a total of 48. Thus the probability for the first question is $frac48120 = frac25$. The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$frac24120 = frac15$. The third question is the two strangers can be picked in $4choose2$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2times6times6) = 72$. Thus the probability is $frac72120 = frac35$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem. Fix yourself, there are 2 ways your known friend can sit either side. The rest of you can sit in 4! ways to a total of 48. Thus the probability for the first question is $frac48120 = frac25$. The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$frac24120 = frac15$. The third question is the two strangers can be picked in $4choose2$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2times6times6) = 72$. Thus the probability is $frac72120 = frac35$






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem. Fix yourself, there are 2 ways your known friend can sit either side. The rest of you can sit in 4! ways to a total of 48. Thus the probability for the first question is $frac48120 = frac25$. The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$frac24120 = frac15$. The third question is the two strangers can be picked in $4choose2$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2times6times6) = 72$. Thus the probability is $frac72120 = frac35$






                    share|cite|improve this answer












                    Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem. Fix yourself, there are 2 ways your known friend can sit either side. The rest of you can sit in 4! ways to a total of 48. Thus the probability for the first question is $frac48120 = frac25$. The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$frac24120 = frac15$. The third question is the two strangers can be picked in $4choose2$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2times6times6) = 72$. Thus the probability is $frac72120 = frac35$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Satish Ramanathan

                    9,17731223




                    9,17731223




















                        up vote
                        0
                        down vote













                        In total, there are $6! = 720$ ways for all the people to sit on the chairs.



                        1. First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$frac 24 times 2 times 6720=frac25$$


                        2. Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$frac 24 times 1 times 6720=frac15$$


                        3. In fact this is the complement of 1. so the probability would be $$1- frac25=frac35$$


                        (Sorry, I was late and English is my second language)






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          In total, there are $6! = 720$ ways for all the people to sit on the chairs.



                          1. First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$frac 24 times 2 times 6720=frac25$$


                          2. Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$frac 24 times 1 times 6720=frac15$$


                          3. In fact this is the complement of 1. so the probability would be $$1- frac25=frac35$$


                          (Sorry, I was late and English is my second language)






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            In total, there are $6! = 720$ ways for all the people to sit on the chairs.



                            1. First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$frac 24 times 2 times 6720=frac25$$


                            2. Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$frac 24 times 1 times 6720=frac15$$


                            3. In fact this is the complement of 1. so the probability would be $$1- frac25=frac35$$


                            (Sorry, I was late and English is my second language)






                            share|cite|improve this answer














                            In total, there are $6! = 720$ ways for all the people to sit on the chairs.



                            1. First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$frac 24 times 2 times 6720=frac25$$


                            2. Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$frac 24 times 1 times 6720=frac15$$


                            3. In fact this is the complement of 1. so the probability would be $$1- frac25=frac35$$


                            (Sorry, I was late and English is my second language)







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








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