Does the specific chair matter in this case?
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
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
add a comment |Â
up vote
-1
down vote
favorite
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
1
I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
â 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)
â Hurkyl
Aug 30 at 4:13
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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
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
probability
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?
â 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)
â Hurkyl
Aug 30 at 4:13
add a comment |Â
1
I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
â 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)
â 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?
â 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?
â 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)
â 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)
â Hurkyl
Aug 30 at 4:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
No, it does not. To the chair, the probability of a boy sitting on it is 7/11.
add a comment |Â
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 $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
No, it does not. To the chair, the probability of a boy sitting on it is 7/11.
add a comment |Â
up vote
0
down vote
No, it does not. To the chair, the probability of a boy sitting on it is 7/11.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
No, it does not. To the chair, the probability of a boy sitting on it is 7/11.
No, it does not. To the chair, the probability of a boy sitting on it is 7/11.
answered Aug 30 at 4:10
Ecotistician
1818
1818
add a comment |Â
add a comment |Â
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 $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$
add a comment |Â
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 $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$
add a comment |Â
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 $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$
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 $$frac7*10!11! =frac 7*(10*9*8*...*2*1)11*(10*9*8*...*2*1) = frac711$$
answered Aug 30 at 4:59
WaveX
2,0411718
2,0411718
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899087%2fdoes-the-specific-chair-matter-in-this-case%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
I presume "at random" means with a uniform distribution. If so, why should the number of the chair matter?
â 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)
â Hurkyl
Aug 30 at 4:13