Mixing
Probability that in one rolling of 5 dice we obtain two '6' and one '5'?
Clash Royale CLAN TAG#URR8PPP
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First I made a mistake of not taking into account that this event is dependent.
So to get two '$6$' the probability would be:
$$ P_5(2) = frac5!2!(5-2)! cdot left(frac16right)^2cdotleft(frac56right)^3 $$
Where
- $p = dfrac16$
- $q = dfrac56$
$N= 5$ (the number of elements of the system)
Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = frac3!1!(3-1)! cdot left(frac16right)^1cdotleft(frac56right)^2 $$
But then the professor told me that $p=dfrac15$ and not $p=dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
probability dice
edited 39 mins ago
Tianlalu
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up vote
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First I made a mistake of not taking into account that this event is dependent.
So to get two '$6$' the probability would be:
$$ P_5(2) = frac5!2!(5-2)! cdot left(frac16right)^2cdotleft(frac56right)^3 $$
Where
- $p = dfrac16$
- $q = dfrac56$
$N= 5$ (the number of elements of the system)
Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = frac3!1!(3-1)! cdot left(frac16right)^1cdotleft(frac56right)^2 $$
But then the professor told me that $p=dfrac15$ and not $p=dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
probability dice
edited 39 mins ago
Tianlalu
1,644620
asked 3 hours ago
Nuz
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Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
First I made a mistake of not taking into account that this event is dependent.
So to get two '$6$' the probability would be:
$$ P_5(2) = frac5!2!(5-2)! cdot left(frac16right)^2cdotleft(frac56right)^3 $$
Where
- $p = dfrac16$
- $q = dfrac56$
$N= 5$ (the number of elements of the system)
Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = frac3!1!(3-1)! cdot left(frac16right)^1cdotleft(frac56right)^2 $$
But then the professor told me that $p=dfrac15$ and not $p=dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
probability dice
edited 39 mins ago
Tianlalu
1,644620
asked 3 hours ago
Nuz
214
New contributor
Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First I made a mistake of not taking into account that this event is dependent.
So to get two '$6$' the probability would be:
$$ P_5(2) = frac5!2!(5-2)! cdot left(frac16right)^2cdotleft(frac56right)^3 $$
Where
- $p = dfrac16$
- $q = dfrac56$
$N= 5$ (the number of elements of the system)
Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = frac3!1!(3-1)! cdot left(frac16right)^1cdotleft(frac56right)^2 $$
But then the professor told me that $p=dfrac15$ and not $p=dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
probability dice
probability dice
edited 39 mins ago
Tianlalu
1,644620
asked 3 hours ago
Nuz
214
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Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 39 mins ago
Tianlalu
1,644620
asked 3 hours ago
Nuz
214
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Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 39 mins ago
Tianlalu
1,644620
edited 39 mins ago
Tianlalu
1,644620
edited 39 mins ago
Tianlalu
1,644620
1,644620
asked 3 hours ago
Nuz
214
New contributor
Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Nuz
214
asked 3 hours ago
Nuz
214
214
New contributor
Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nuz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago
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2
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago
2
2
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago
add a comment |Â
2 Answers
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oldest
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up vote
3
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I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_other = 2) = frac5!2!1!2! left( frac16 right)^2 left( frac16 right) left( frac46 right)^2 approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F mid E)P(G mid EF) $$
$$P(E) = 5 choose 2left( frac16 right)^2 left( frac56 right)^3 $$
$$P(F mid E) = 3 choose 1left( frac15 right) left( frac45 right)^2$$
$$P(G mid EF) = 2 choose 2left( frac44 right)^2 left( frac44 right)^0 = 1$$
$$P(EFG) approx .062$$
answered 3 hours ago
HJ_beginner
787215
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
add a comment |Â
up vote
2
down vote
But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$mathsf P(N_6=2)=binom 52 dfrac1^25^36^5$$
Now, of the three dice which don't show 6, each may show faces $1,2,3,4,5$ with equal probability. Â Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$. Â So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$mathsf P(N_5=1mid N_6=2)=binom 31dfrac1^14^25^3$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $5^3$ factors will cancel, so:
$$mathsf P(N_6=2,N_5=1)=binom52binom31dfrac1^21^14^26^5$$
PS: This is called a multinomial distribution.
answered 22 mins ago
Graham Kemp
82.2k43377
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_other = 2) = frac5!2!1!2! left( frac16 right)^2 left( frac16 right) left( frac46 right)^2 approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F mid E)P(G mid EF) $$
$$P(E) = 5 choose 2left( frac16 right)^2 left( frac56 right)^3 $$
$$P(F mid E) = 3 choose 1left( frac15 right) left( frac45 right)^2$$
$$P(G mid EF) = 2 choose 2left( frac44 right)^2 left( frac44 right)^0 = 1$$
$$P(EFG) approx .062$$
answered 3 hours ago
HJ_beginner
787215
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
add a comment |Â
up vote
3
down vote
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_other = 2) = frac5!2!1!2! left( frac16 right)^2 left( frac16 right) left( frac46 right)^2 approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F mid E)P(G mid EF) $$
$$P(E) = 5 choose 2left( frac16 right)^2 left( frac56 right)^3 $$
$$P(F mid E) = 3 choose 1left( frac15 right) left( frac45 right)^2$$
$$P(G mid EF) = 2 choose 2left( frac44 right)^2 left( frac44 right)^0 = 1$$
$$P(EFG) approx .062$$
answered 3 hours ago
HJ_beginner
787215
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_other = 2) = frac5!2!1!2! left( frac16 right)^2 left( frac16 right) left( frac46 right)^2 approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F mid E)P(G mid EF) $$
$$P(E) = 5 choose 2left( frac16 right)^2 left( frac56 right)^3 $$
$$P(F mid E) = 3 choose 1left( frac15 right) left( frac45 right)^2$$
$$P(G mid EF) = 2 choose 2left( frac44 right)^2 left( frac44 right)^0 = 1$$
$$P(EFG) approx .062$$
answered 3 hours ago
HJ_beginner
787215
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_other = 2) = frac5!2!1!2! left( frac16 right)^2 left( frac16 right) left( frac46 right)^2 approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F mid E)P(G mid EF) $$
$$P(E) = 5 choose 2left( frac16 right)^2 left( frac56 right)^3 $$
$$P(F mid E) = 3 choose 1left( frac15 right) left( frac45 right)^2$$
$$P(G mid EF) = 2 choose 2left( frac44 right)^2 left( frac44 right)^0 = 1$$
$$P(EFG) approx .062$$
answered 3 hours ago
HJ_beginner
787215
answered 3 hours ago
HJ_beginner
787215
answered 3 hours ago
HJ_beginner
787215
answered 3 hours ago
HJ_beginner
787215
787215
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
add a comment |Â
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces.
– Nuz
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space.
– HJ_beginner
3 hours ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something?
– Nuz
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F mid G)P(E mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting.
– HJ_beginner
1 hour ago
add a comment |Â
up vote
2
down vote
But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$mathsf P(N_6=2)=binom 52 dfrac1^25^36^5$$
Now, of the three dice which don't show 6, each may show faces $1,2,3,4,5$ with equal probability. Â Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$. Â So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$mathsf P(N_5=1mid N_6=2)=binom 31dfrac1^14^25^3$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $5^3$ factors will cancel, so:
$$mathsf P(N_6=2,N_5=1)=binom52binom31dfrac1^21^14^26^5$$
PS: This is called a multinomial distribution.
answered 22 mins ago
Graham Kemp
82.2k43377
add a comment |Â
up vote
2
down vote
But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$mathsf P(N_6=2)=binom 52 dfrac1^25^36^5$$
Now, of the three dice which don't show 6, each may show faces $1,2,3,4,5$ with equal probability. Â Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$. Â So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$mathsf P(N_5=1mid N_6=2)=binom 31dfrac1^14^25^3$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $5^3$ factors will cancel, so:
$$mathsf P(N_6=2,N_5=1)=binom52binom31dfrac1^21^14^26^5$$
PS: This is called a multinomial distribution.
answered 22 mins ago
Graham Kemp
82.2k43377
add a comment |Â
up vote
2
down vote
up vote
2
down vote
But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$mathsf P(N_6=2)=binom 52 dfrac1^25^36^5$$
Now, of the three dice which don't show 6, each may show faces $1,2,3,4,5$ with equal probability. Â Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$. Â So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$mathsf P(N_5=1mid N_6=2)=binom 31dfrac1^14^25^3$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $5^3$ factors will cancel, so:
$$mathsf P(N_6=2,N_5=1)=binom52binom31dfrac1^21^14^26^5$$
PS: This is called a multinomial distribution.
answered 22 mins ago
Graham Kemp
82.2k43377
But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$mathsf P(N_6=2)=binom 52 dfrac1^25^36^5$$
Now, of the three dice which don't show 6, each may show faces $1,2,3,4,5$ with equal probability. Â Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$. Â So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$mathsf P(N_5=1mid N_6=2)=binom 31dfrac1^14^25^3$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $5^3$ factors will cancel, so:
$$mathsf P(N_6=2,N_5=1)=binom52binom31dfrac1^21^14^26^5$$
PS: This is called a multinomial distribution.
answered 22 mins ago
Graham Kemp
82.2k43377
answered 22 mins ago
Graham Kemp
82.2k43377
answered 22 mins ago
Graham Kemp
82.2k43377
answered 22 mins ago
Graham Kemp
82.2k43377
82.2k43377
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2
The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s.
– StatGuy
3 hours ago