Mixing
9 person randomly enter 3 different rooms
$begingroup$
9 person randomly enter 3 different rooms. What is the probability that
a)the first room has 3 person?
b)every room has 3 person?
c)the first room has n person, second room has 3 persons, third room 2 persons?
What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only.
probability
asked Jan 27 at 9:26
CColaCCola
376
$endgroup$
add a comment |
$begingroup$
9 person randomly enter 3 different rooms. What is the probability that
a)the first room has 3 person?
b)every room has 3 person?
c)the first room has n person, second room has 3 persons, third room 2 persons?
What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only.
probability
asked Jan 27 at 9:26
CColaCCola
376
$endgroup$
add a comment |
$begingroup$
9 person randomly enter 3 different rooms. What is the probability that
a)the first room has 3 person?
b)every room has 3 person?
c)the first room has n person, second room has 3 persons, third room 2 persons?
What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only.
probability
asked Jan 27 at 9:26
CColaCCola
376
$endgroup$
9 person randomly enter 3 different rooms. What is the probability that
a)the first room has 3 person?
b)every room has 3 person?
c)the first room has n person, second room has 3 persons, third room 2 persons?
What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only.
probability
probability
asked Jan 27 at 9:26
CColaCCola
376
asked Jan 27 at 9:26
CColaCCola
376
asked Jan 27 at 9:26
CColaCCola
376
asked Jan 27 at 9:26
CColaCCola
376
asked Jan 27 at 9:26
CColaCCola
376
376
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
"$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
eqalign{
& left( {r_{,1} + r_{,2} + r_{,3} } right)^{,9} = cr
& = cdots + r_{j_{,1} } ,r_{j_{,2} } ,r_{j_{,3} } , cdots r_{j_{,8} } ,r_{j_{,9} } ,
+ cdots quad left| {;j_{,k} in left{ {1,2,3} right}} right. = cr
& = sumlimits_{k_{,1} + k_{,2} + k_{,3} ; = ,9}
{;left( matrix{ 9 cr k_{,1} ,k_{,2} ,k_{,3} cr} right)r_{,1} ^{,k_{,1} } r_{,2} ^{,k_{,2} } r_{,3} ^{,k_{,3} } } cr}
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is
$$
eqalign{
& sumlimits_{k_{,2} + k_{,3} ; = ,6} {;left( matrix{ 9 cr 3,k_{,2} ,k_{,3} cr} right)}
= sumlimits_{k_{,2} + k_{,3} ; = ,6} {;{{9!} over {3!k_{,2} !k_{,3} !}}} = cr
& = left( matrix{ 9 hfill cr 3 hfill cr} right)sumlimits_k
{;left( matrix{ 6 hfill cr k hfill cr} right)}
= left( matrix{ 9 hfill cr 3 hfill cr} right)2^{,6} cr}
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
left( {n_{,1} ,n_{,2} ,n_{,3} } right)quad left| matrix{
;0 le n_{,k} le 9 hfill cr
;n_{,1} + n_{,2} + n_{,3} = 9 hfill cr} right.
$$
is considered equi-probable.
There are
$$
binom{9+3-1}{3-1} = binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which
$$
binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
add a comment |
$begingroup$
a) $binom93*2^6=5376$
b) $binom93*binom63=1680$
c) $binom94*binom53=1260$
answered Jan 27 at 9:46
abc...abc...
3,237738
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
"$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
eqalign{
& left( {r_{,1} + r_{,2} + r_{,3} } right)^{,9} = cr
& = cdots + r_{j_{,1} } ,r_{j_{,2} } ,r_{j_{,3} } , cdots r_{j_{,8} } ,r_{j_{,9} } ,
+ cdots quad left| {;j_{,k} in left{ {1,2,3} right}} right. = cr
& = sumlimits_{k_{,1} + k_{,2} + k_{,3} ; = ,9}
{;left( matrix{ 9 cr k_{,1} ,k_{,2} ,k_{,3} cr} right)r_{,1} ^{,k_{,1} } r_{,2} ^{,k_{,2} } r_{,3} ^{,k_{,3} } } cr}
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is
$$
eqalign{
& sumlimits_{k_{,2} + k_{,3} ; = ,6} {;left( matrix{ 9 cr 3,k_{,2} ,k_{,3} cr} right)}
= sumlimits_{k_{,2} + k_{,3} ; = ,6} {;{{9!} over {3!k_{,2} !k_{,3} !}}} = cr
& = left( matrix{ 9 hfill cr 3 hfill cr} right)sumlimits_k
{;left( matrix{ 6 hfill cr k hfill cr} right)}
= left( matrix{ 9 hfill cr 3 hfill cr} right)2^{,6} cr}
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
left( {n_{,1} ,n_{,2} ,n_{,3} } right)quad left| matrix{
;0 le n_{,k} le 9 hfill cr
;n_{,1} + n_{,2} + n_{,3} = 9 hfill cr} right.
$$
is considered equi-probable.
There are
$$
binom{9+3-1}{3-1} = binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which
$$
binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
add a comment |
$begingroup$
"$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
eqalign{
& left( {r_{,1} + r_{,2} + r_{,3} } right)^{,9} = cr
& = cdots + r_{j_{,1} } ,r_{j_{,2} } ,r_{j_{,3} } , cdots r_{j_{,8} } ,r_{j_{,9} } ,
+ cdots quad left| {;j_{,k} in left{ {1,2,3} right}} right. = cr
& = sumlimits_{k_{,1} + k_{,2} + k_{,3} ; = ,9}
{;left( matrix{ 9 cr k_{,1} ,k_{,2} ,k_{,3} cr} right)r_{,1} ^{,k_{,1} } r_{,2} ^{,k_{,2} } r_{,3} ^{,k_{,3} } } cr}
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is
$$
eqalign{
& sumlimits_{k_{,2} + k_{,3} ; = ,6} {;left( matrix{ 9 cr 3,k_{,2} ,k_{,3} cr} right)}
= sumlimits_{k_{,2} + k_{,3} ; = ,6} {;{{9!} over {3!k_{,2} !k_{,3} !}}} = cr
& = left( matrix{ 9 hfill cr 3 hfill cr} right)sumlimits_k
{;left( matrix{ 6 hfill cr k hfill cr} right)}
= left( matrix{ 9 hfill cr 3 hfill cr} right)2^{,6} cr}
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
left( {n_{,1} ,n_{,2} ,n_{,3} } right)quad left| matrix{
;0 le n_{,k} le 9 hfill cr
;n_{,1} + n_{,2} + n_{,3} = 9 hfill cr} right.
$$
is considered equi-probable.
There are
$$
binom{9+3-1}{3-1} = binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which
$$
binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
add a comment |
$begingroup$
"$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
eqalign{
& left( {r_{,1} + r_{,2} + r_{,3} } right)^{,9} = cr
& = cdots + r_{j_{,1} } ,r_{j_{,2} } ,r_{j_{,3} } , cdots r_{j_{,8} } ,r_{j_{,9} } ,
+ cdots quad left| {;j_{,k} in left{ {1,2,3} right}} right. = cr
& = sumlimits_{k_{,1} + k_{,2} + k_{,3} ; = ,9}
{;left( matrix{ 9 cr k_{,1} ,k_{,2} ,k_{,3} cr} right)r_{,1} ^{,k_{,1} } r_{,2} ^{,k_{,2} } r_{,3} ^{,k_{,3} } } cr}
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is
$$
eqalign{
& sumlimits_{k_{,2} + k_{,3} ; = ,6} {;left( matrix{ 9 cr 3,k_{,2} ,k_{,3} cr} right)}
= sumlimits_{k_{,2} + k_{,3} ; = ,6} {;{{9!} over {3!k_{,2} !k_{,3} !}}} = cr
& = left( matrix{ 9 hfill cr 3 hfill cr} right)sumlimits_k
{;left( matrix{ 6 hfill cr k hfill cr} right)}
= left( matrix{ 9 hfill cr 3 hfill cr} right)2^{,6} cr}
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
left( {n_{,1} ,n_{,2} ,n_{,3} } right)quad left| matrix{
;0 le n_{,k} le 9 hfill cr
;n_{,1} + n_{,2} + n_{,3} = 9 hfill cr} right.
$$
is considered equi-probable.
There are
$$
binom{9+3-1}{3-1} = binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which
$$
binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
$endgroup$
"$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
eqalign{
& left( {r_{,1} + r_{,2} + r_{,3} } right)^{,9} = cr
& = cdots + r_{j_{,1} } ,r_{j_{,2} } ,r_{j_{,3} } , cdots r_{j_{,8} } ,r_{j_{,9} } ,
+ cdots quad left| {;j_{,k} in left{ {1,2,3} right}} right. = cr
& = sumlimits_{k_{,1} + k_{,2} + k_{,3} ; = ,9}
{;left( matrix{ 9 cr k_{,1} ,k_{,2} ,k_{,3} cr} right)r_{,1} ^{,k_{,1} } r_{,2} ^{,k_{,2} } r_{,3} ^{,k_{,3} } } cr}
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is
$$
eqalign{
& sumlimits_{k_{,2} + k_{,3} ; = ,6} {;left( matrix{ 9 cr 3,k_{,2} ,k_{,3} cr} right)}
= sumlimits_{k_{,2} + k_{,3} ; = ,6} {;{{9!} over {3!k_{,2} !k_{,3} !}}} = cr
& = left( matrix{ 9 hfill cr 3 hfill cr} right)sumlimits_k
{;left( matrix{ 6 hfill cr k hfill cr} right)}
= left( matrix{ 9 hfill cr 3 hfill cr} right)2^{,6} cr}
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
left( {n_{,1} ,n_{,2} ,n_{,3} } right)quad left| matrix{
;0 le n_{,k} le 9 hfill cr
;n_{,1} + n_{,2} + n_{,3} = 9 hfill cr} right.
$$
is considered equi-probable.
There are
$$
binom{9+3-1}{3-1} = binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which
$$
binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
edited Jan 27 at 18:09
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
answered Jan 27 at 9:33
G CabG Cab
20.4k31341
20.4k31341
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
add a comment |
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
If the people are distinguishable, then it's not stars and bars. If the people are identical, then the stars and bars (distinguishable) outcomes are not equally likely (assuming "randomly" means "uniformly and independent" as it usually does), which makes it a hard way to answer the probability questions here.
$endgroup$
– Ned
Jan 27 at 15:51
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
$begingroup$
@Ned: you are totally right ! I answered too hurriedly.. I amended my answer according to possible interpretation of the random process of room occupation.
$endgroup$
– G Cab
Jan 27 at 18:11
add a comment |
$begingroup$
a) $binom93*2^6=5376$
b) $binom93*binom63=1680$
c) $binom94*binom53=1260$
answered Jan 27 at 9:46
abc...abc...
3,237738
$endgroup$
add a comment |
$begingroup$
a) $binom93*2^6=5376$
b) $binom93*binom63=1680$
c) $binom94*binom53=1260$
answered Jan 27 at 9:46
abc...abc...
3,237738
$endgroup$
add a comment |
$begingroup$
a) $binom93*2^6=5376$
b) $binom93*binom63=1680$
c) $binom94*binom53=1260$
answered Jan 27 at 9:46
abc...abc...
3,237738
$endgroup$
a) $binom93*2^6=5376$
b) $binom93*binom63=1680$
c) $binom94*binom53=1260$
answered Jan 27 at 9:46
abc...abc...
3,237738
answered Jan 27 at 9:46
abc...abc...
3,237738
answered Jan 27 at 9:46
abc...abc...
3,237738
answered Jan 27 at 9:46
abc...abc...
3,237738
3,237738
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
