Mixing
Twelve fair dice are rolled at random. Calculate the probability that each number 1, 2, 3, 4, 5, 6 appears...
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Ok so my effort:
(1/6)^12 is the probability of getting exactly 1,1,2,2,3,3,4,4,5,5,6,6 in that order
I know you need to consider using choose, but I have no idea which one to use here.
12Choose?? .
probability
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
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add a comment |
$begingroup$
Ok so my effort:
(1/6)^12 is the probability of getting exactly 1,1,2,2,3,3,4,4,5,5,6,6 in that order
I know you need to consider using choose, but I have no idea which one to use here.
12Choose?? .
probability
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
$endgroup$
add a comment |
$begingroup$
Ok so my effort:
(1/6)^12 is the probability of getting exactly 1,1,2,2,3,3,4,4,5,5,6,6 in that order
I know you need to consider using choose, but I have no idea which one to use here.
12Choose?? .
probability
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
$endgroup$
Ok so my effort:
(1/6)^12 is the probability of getting exactly 1,1,2,2,3,3,4,4,5,5,6,6 in that order
I know you need to consider using choose, but I have no idea which one to use here.
12Choose?? .
probability
probability
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
asked Jan 23 at 16:07
AnoUser1AnoUser1
836
836
add a comment |
add a comment |
3 Answers
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$begingroup$
Your effort is okay, and you should wonder now: "how many orders exist such that every number in ${1,dots,6}$ occurs twice?"
The answer on that is $$binom{12}{2,2,2,2,2,2}:=frac{12!}{2!2!2!2!2!2!}$$
You are dealing here with multinomial distribution.
There are $12$ independent experiments that have $6$ equiprobable outcomes (in stead of $2$ as in the binomial case).
The probability you mention equals:$$6^{-12}frac{12!}{2!2!2!2!2!2!}$$
answered Jan 23 at 16:30
drhabdrhab
103k545136
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add a comment |
$begingroup$
You have to multiply your answer by the number of rearrangements of 1,1,2,2,3,3,4,4,5,5,6,6 to finish. That is $12!/(2^6)$ using multinomial coefficient.
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
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add a comment |
$begingroup$
Good so far. Now if the string was $1,2,cdots, 12$ you could permute it in $12!$ ways, while applying that to a string with $6 couples, each couple will be counted twice. So ..
answered Jan 23 at 16:33
G CabG Cab
20k31340
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3 Answers
3
active
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3 Answers
3
active
oldest
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active
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active
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$begingroup$
Your effort is okay, and you should wonder now: "how many orders exist such that every number in ${1,dots,6}$ occurs twice?"
The answer on that is $$binom{12}{2,2,2,2,2,2}:=frac{12!}{2!2!2!2!2!2!}$$
You are dealing here with multinomial distribution.
There are $12$ independent experiments that have $6$ equiprobable outcomes (in stead of $2$ as in the binomial case).
The probability you mention equals:$$6^{-12}frac{12!}{2!2!2!2!2!2!}$$
answered Jan 23 at 16:30
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Your effort is okay, and you should wonder now: "how many orders exist such that every number in ${1,dots,6}$ occurs twice?"
The answer on that is $$binom{12}{2,2,2,2,2,2}:=frac{12!}{2!2!2!2!2!2!}$$
You are dealing here with multinomial distribution.
There are $12$ independent experiments that have $6$ equiprobable outcomes (in stead of $2$ as in the binomial case).
The probability you mention equals:$$6^{-12}frac{12!}{2!2!2!2!2!2!}$$
answered Jan 23 at 16:30
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Your effort is okay, and you should wonder now: "how many orders exist such that every number in ${1,dots,6}$ occurs twice?"
The answer on that is $$binom{12}{2,2,2,2,2,2}:=frac{12!}{2!2!2!2!2!2!}$$
You are dealing here with multinomial distribution.
There are $12$ independent experiments that have $6$ equiprobable outcomes (in stead of $2$ as in the binomial case).
The probability you mention equals:$$6^{-12}frac{12!}{2!2!2!2!2!2!}$$
answered Jan 23 at 16:30
drhabdrhab
103k545136
$endgroup$
Your effort is okay, and you should wonder now: "how many orders exist such that every number in ${1,dots,6}$ occurs twice?"
The answer on that is $$binom{12}{2,2,2,2,2,2}:=frac{12!}{2!2!2!2!2!2!}$$
You are dealing here with multinomial distribution.
There are $12$ independent experiments that have $6$ equiprobable outcomes (in stead of $2$ as in the binomial case).
The probability you mention equals:$$6^{-12}frac{12!}{2!2!2!2!2!2!}$$
answered Jan 23 at 16:30
drhabdrhab
103k545136
answered Jan 23 at 16:30
drhabdrhab
103k545136
answered Jan 23 at 16:30
drhabdrhab
103k545136
answered Jan 23 at 16:30
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
You have to multiply your answer by the number of rearrangements of 1,1,2,2,3,3,4,4,5,5,6,6 to finish. That is $12!/(2^6)$ using multinomial coefficient.
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
$endgroup$
add a comment |
$begingroup$
You have to multiply your answer by the number of rearrangements of 1,1,2,2,3,3,4,4,5,5,6,6 to finish. That is $12!/(2^6)$ using multinomial coefficient.
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
$endgroup$
add a comment |
$begingroup$
You have to multiply your answer by the number of rearrangements of 1,1,2,2,3,3,4,4,5,5,6,6 to finish. That is $12!/(2^6)$ using multinomial coefficient.
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
$endgroup$
You have to multiply your answer by the number of rearrangements of 1,1,2,2,3,3,4,4,5,5,6,6 to finish. That is $12!/(2^6)$ using multinomial coefficient.
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
answered Jan 23 at 16:29
coffeemathcoffeemath
2,8971415
2,8971415
add a comment |
add a comment |
$begingroup$
Good so far. Now if the string was $1,2,cdots, 12$ you could permute it in $12!$ ways, while applying that to a string with $6 couples, each couple will be counted twice. So ..
answered Jan 23 at 16:33
G CabG Cab
20k31340
$endgroup$
add a comment |
$begingroup$
Good so far. Now if the string was $1,2,cdots, 12$ you could permute it in $12!$ ways, while applying that to a string with $6 couples, each couple will be counted twice. So ..
answered Jan 23 at 16:33
G CabG Cab
20k31340
$endgroup$
add a comment |
$begingroup$
Good so far. Now if the string was $1,2,cdots, 12$ you could permute it in $12!$ ways, while applying that to a string with $6 couples, each couple will be counted twice. So ..
answered Jan 23 at 16:33
G CabG Cab
20k31340
$endgroup$
Good so far. Now if the string was $1,2,cdots, 12$ you could permute it in $12!$ ways, while applying that to a string with $6 couples, each couple will be counted twice. So ..
answered Jan 23 at 16:33
G CabG Cab
20k31340
answered Jan 23 at 16:33
G CabG Cab
20k31340
answered Jan 23 at 16:33
G CabG Cab
20k31340
answered Jan 23 at 16:33
G CabG Cab
20k31340
20k31340
add a comment |
add a comment |
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