Does the specific chair matter in this case?

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Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these
chairs at random. What is the probability that chair 5 is occupied by a boy? Is it 7/11 or does the fact that the chair number is specified play a role?










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    I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
    – Lord Shark the Unknown
    Aug 30 at 4:03











  • @LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
    – Hurkyl
    Aug 30 at 4:13















up vote
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Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these
chairs at random. What is the probability that chair 5 is occupied by a boy? Is it 7/11 or does the fact that the chair number is specified play a role?










share|cite|improve this question

















  • 1




    I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
    – Lord Shark the Unknown
    Aug 30 at 4:03











  • @LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
    – Hurkyl
    Aug 30 at 4:13













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Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these
chairs at random. What is the probability that chair 5 is occupied by a boy? Is it 7/11 or does the fact that the chair number is specified play a role?










share|cite|improve this question













Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these
chairs at random. What is the probability that chair 5 is occupied by a boy? Is it 7/11 or does the fact that the chair number is specified play a role?







probability






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asked Aug 30 at 3:39









Murph Jones

502




502







  • 1




    I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
    – Lord Shark the Unknown
    Aug 30 at 4:03











  • @LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
    – Hurkyl
    Aug 30 at 4:13













  • 1




    I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
    – Lord Shark the Unknown
    Aug 30 at 4:03











  • @LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
    – Hurkyl
    Aug 30 at 4:13








1




1




I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
– Lord Shark the Unknown
Aug 30 at 4:03





I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
– Lord Shark the Unknown
Aug 30 at 4:03













@LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
– Hurkyl
Aug 30 at 4:13





@LordSharktheUnknown: Because added information can have all sorts of subtle effects that trip up people who brush off details like this. How is a student to know that this happens to be one of the instances where the seemingly obvious method is actually correct? (that's not a rhetorical question -- a way to distinguish which 'obvious' simplifications are right and which are mistakes is something one needs to efficiently and correctly solve probability problems)
– Hurkyl
Aug 30 at 4:13











2 Answers
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No, it does not. To the chair, the probability of a boy sitting on it is 7/11.






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    Here's a way to see why you are correct with $7/11$ and the chairs being numbered don't matter. Let's decide who sits in chair $5$ first then decide the rest of the chairs afterwards. We have $7$ guys to choose for that particular chair, then we have $10$ remaining people for the next chair, $9$ for the next, and so on, so our numerator is $7*10!$



    The total number of ways to seat them is $11!$



    So, simplifying, the probably a guy sits in seat $5$ is $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$






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      2 Answers
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      No, it does not. To the chair, the probability of a boy sitting on it is 7/11.






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        No, it does not. To the chair, the probability of a boy sitting on it is 7/11.






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          No, it does not. To the chair, the probability of a boy sitting on it is 7/11.






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          No, it does not. To the chair, the probability of a boy sitting on it is 7/11.







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          answered Aug 30 at 4:10









          Ecotistician

          1818




          1818




















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              down vote













              Here's a way to see why you are correct with $7/11$ and the chairs being numbered don't matter. Let's decide who sits in chair $5$ first then decide the rest of the chairs afterwards. We have $7$ guys to choose for that particular chair, then we have $10$ remaining people for the next chair, $9$ for the next, and so on, so our numerator is $7*10!$



              The total number of ways to seat them is $11!$



              So, simplifying, the probably a guy sits in seat $5$ is $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$






              share|cite|improve this answer
























                up vote
                0
                down vote













                Here's a way to see why you are correct with $7/11$ and the chairs being numbered don't matter. Let's decide who sits in chair $5$ first then decide the rest of the chairs afterwards. We have $7$ guys to choose for that particular chair, then we have $10$ remaining people for the next chair, $9$ for the next, and so on, so our numerator is $7*10!$



                The total number of ways to seat them is $11!$



                So, simplifying, the probably a guy sits in seat $5$ is $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Here's a way to see why you are correct with $7/11$ and the chairs being numbered don't matter. Let's decide who sits in chair $5$ first then decide the rest of the chairs afterwards. We have $7$ guys to choose for that particular chair, then we have $10$ remaining people for the next chair, $9$ for the next, and so on, so our numerator is $7*10!$



                  The total number of ways to seat them is $11!$



                  So, simplifying, the probably a guy sits in seat $5$ is $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$






                  share|cite|improve this answer












                  Here's a way to see why you are correct with $7/11$ and the chairs being numbered don't matter. Let's decide who sits in chair $5$ first then decide the rest of the chairs afterwards. We have $7$ guys to choose for that particular chair, then we have $10$ remaining people for the next chair, $9$ for the next, and so on, so our numerator is $7*10!$



                  The total number of ways to seat them is $11!$



                  So, simplifying, the probably a guy sits in seat $5$ is $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 30 at 4:59









                  WaveX

                  2,0411718




                  2,0411718



























                       

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