Mixing
Probability of 2 Dice Throws Equal to Sum
Clash Royale CLAN TAG#URR8PPP
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I'm aware this is a very simple question, I must be missing something obvious.
Given that the numbers coming out of two independent dice throws are different, find the probability that that sum of the numbers is 4.
Apparently the correct answer is 1/15. Though my answer is 1/18:
So you can roll either a $(1,3)$ or a $(3,1)$. So then I do ($frac16cdotfrac16$) + ($frac16cdotfrac16$) = $frac118$
probability dice
asked 3 hours ago
the-realtom
1135
add a comment |Â
up vote
2
down vote
favorite
I'm aware this is a very simple question, I must be missing something obvious.
Given that the numbers coming out of two independent dice throws are different, find the probability that that sum of the numbers is 4.
Apparently the correct answer is 1/15. Though my answer is 1/18:
So you can roll either a $(1,3)$ or a $(3,1)$. So then I do ($frac16cdotfrac16$) + ($frac16cdotfrac16$) = $frac118$
probability dice
asked 3 hours ago
the-realtom
1135
2
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm aware this is a very simple question, I must be missing something obvious.
Given that the numbers coming out of two independent dice throws are different, find the probability that that sum of the numbers is 4.
Apparently the correct answer is 1/15. Though my answer is 1/18:
So you can roll either a $(1,3)$ or a $(3,1)$. So then I do ($frac16cdotfrac16$) + ($frac16cdotfrac16$) = $frac118$
probability dice
asked 3 hours ago
the-realtom
1135
I'm aware this is a very simple question, I must be missing something obvious.
Given that the numbers coming out of two independent dice throws are different, find the probability that that sum of the numbers is 4.
Apparently the correct answer is 1/15. Though my answer is 1/18:
So you can roll either a $(1,3)$ or a $(3,1)$. So then I do ($frac16cdotfrac16$) + ($frac16cdotfrac16$) = $frac118$
probability dice
probability dice
asked 3 hours ago
the-realtom
1135
asked 3 hours ago
the-realtom
1135
asked 3 hours ago
the-realtom
1135
asked 3 hours ago
the-realtom
1135
asked 3 hours ago
the-realtom
1135
1135
2
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago
add a comment |Â
2
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago
2
2
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
One way to approach this problem is to count (as you did) the total number of rolls that could give you 4, and then divide by the total number of possible rolls where no two dice are the same.
As you mentioned the number of rolls that give you $4$ is $2$, i.e. $(3,1)$ and $(1,3)$.
However, I think why you're getting $frac118$ is because you then divide by all possible rolls that could occur which is $36$, so you get $frac236$. Instead, you have to divide by all allowable rolls, i.e. rolls that do not have two of the same number. There are $30$ such rolls. To see this you remove all the pairs $(1,1),(2,2),$ etc. from the $36$ total rolls.
This gives you the probability you're looking for:
$frac230 = frac115$
In general, to think of such probability problems you count the amount of what you're looking for and divide by all allowable outcomes you choose from.
answered 3 hours ago
rpl19
2211
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
add a comment |Â
up vote
4
down vote
You are given that the numbers on the dice are different, so there are only $30$ throws. We delete the $6$ doubles from the usual $36$. There are two successes, so the chance is $frac 230=frac 115$
answered 3 hours ago
Ross Millikan
283k23191359
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
One way to approach this problem is to count (as you did) the total number of rolls that could give you 4, and then divide by the total number of possible rolls where no two dice are the same.
As you mentioned the number of rolls that give you $4$ is $2$, i.e. $(3,1)$ and $(1,3)$.
However, I think why you're getting $frac118$ is because you then divide by all possible rolls that could occur which is $36$, so you get $frac236$. Instead, you have to divide by all allowable rolls, i.e. rolls that do not have two of the same number. There are $30$ such rolls. To see this you remove all the pairs $(1,1),(2,2),$ etc. from the $36$ total rolls.
This gives you the probability you're looking for:
$frac230 = frac115$
In general, to think of such probability problems you count the amount of what you're looking for and divide by all allowable outcomes you choose from.
answered 3 hours ago
rpl19
2211
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
One way to approach this problem is to count (as you did) the total number of rolls that could give you 4, and then divide by the total number of possible rolls where no two dice are the same.
As you mentioned the number of rolls that give you $4$ is $2$, i.e. $(3,1)$ and $(1,3)$.
However, I think why you're getting $frac118$ is because you then divide by all possible rolls that could occur which is $36$, so you get $frac236$. Instead, you have to divide by all allowable rolls, i.e. rolls that do not have two of the same number. There are $30$ such rolls. To see this you remove all the pairs $(1,1),(2,2),$ etc. from the $36$ total rolls.
This gives you the probability you're looking for:
$frac230 = frac115$
In general, to think of such probability problems you count the amount of what you're looking for and divide by all allowable outcomes you choose from.
answered 3 hours ago
rpl19
2211
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
One way to approach this problem is to count (as you did) the total number of rolls that could give you 4, and then divide by the total number of possible rolls where no two dice are the same.
As you mentioned the number of rolls that give you $4$ is $2$, i.e. $(3,1)$ and $(1,3)$.
However, I think why you're getting $frac118$ is because you then divide by all possible rolls that could occur which is $36$, so you get $frac236$. Instead, you have to divide by all allowable rolls, i.e. rolls that do not have two of the same number. There are $30$ such rolls. To see this you remove all the pairs $(1,1),(2,2),$ etc. from the $36$ total rolls.
This gives you the probability you're looking for:
$frac230 = frac115$
In general, to think of such probability problems you count the amount of what you're looking for and divide by all allowable outcomes you choose from.
answered 3 hours ago
rpl19
2211
One way to approach this problem is to count (as you did) the total number of rolls that could give you 4, and then divide by the total number of possible rolls where no two dice are the same.
As you mentioned the number of rolls that give you $4$ is $2$, i.e. $(3,1)$ and $(1,3)$.
However, I think why you're getting $frac118$ is because you then divide by all possible rolls that could occur which is $36$, so you get $frac236$. Instead, you have to divide by all allowable rolls, i.e. rolls that do not have two of the same number. There are $30$ such rolls. To see this you remove all the pairs $(1,1),(2,2),$ etc. from the $36$ total rolls.
This gives you the probability you're looking for:
$frac230 = frac115$
In general, to think of such probability problems you count the amount of what you're looking for and divide by all allowable outcomes you choose from.
answered 3 hours ago
rpl19
2211
answered 3 hours ago
rpl19
2211
answered 3 hours ago
rpl19
2211
answered 3 hours ago
rpl19
2211
2211
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
add a comment |Â
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
Ahh okay thanks. Yeah I guess I interpreted the question wrong, I just removed the (2,2) rolls from my count instead of looking at it in context of a sample space without any double combinations.
– the-realtom
3 hours ago
add a comment |Â
up vote
4
down vote
You are given that the numbers on the dice are different, so there are only $30$ throws. We delete the $6$ doubles from the usual $36$. There are two successes, so the chance is $frac 230=frac 115$
answered 3 hours ago
Ross Millikan
283k23191359
add a comment |Â
up vote
4
down vote
You are given that the numbers on the dice are different, so there are only $30$ throws. We delete the $6$ doubles from the usual $36$. There are two successes, so the chance is $frac 230=frac 115$
answered 3 hours ago
Ross Millikan
283k23191359
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You are given that the numbers on the dice are different, so there are only $30$ throws. We delete the $6$ doubles from the usual $36$. There are two successes, so the chance is $frac 230=frac 115$
answered 3 hours ago
Ross Millikan
283k23191359
You are given that the numbers on the dice are different, so there are only $30$ throws. We delete the $6$ doubles from the usual $36$. There are two successes, so the chance is $frac 230=frac 115$
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
283k23191359
add a comment |Â
add a comment |Â
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2
This is conditional probability because your sample space is not all possible throws but only those throws where the two dice are showing different outcomes.
– Anurag A
3 hours ago