Probability of 2 Dice Throws Equal to Sum
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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 the 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 $left(frac16cdotfrac16right)$ + $left(frac16cdotfrac16right)$ = $frac118$
probability dice
edited Oct 4 at 8:32
pointguard0
1,248821
asked Oct 4 at 4:38
the-realtom
1185
add a comment |Â
up vote
3
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 the 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 $left(frac16cdotfrac16right)$ + $left(frac16cdotfrac16right)$ = $frac118$
probability dice
edited Oct 4 at 8:32
pointguard0
1,248821
asked Oct 4 at 4:38
the-realtom
1185
3
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
Oct 4 at 4:41
add a comment |Â
up vote
3
down vote
favorite
up vote
3
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 the 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 $left(frac16cdotfrac16right)$ + $left(frac16cdotfrac16right)$ = $frac118$
probability dice
edited Oct 4 at 8:32
pointguard0
1,248821
asked Oct 4 at 4:38
the-realtom
1185
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 the 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 $left(frac16cdotfrac16right)$ + $left(frac16cdotfrac16right)$ = $frac118$
probability dice
probability dice
edited Oct 4 at 8:32
pointguard0
1,248821
asked Oct 4 at 4:38
the-realtom
1185
edited Oct 4 at 8:32
pointguard0
1,248821
asked Oct 4 at 4:38
the-realtom
1185
edited Oct 4 at 8:32
pointguard0
1,248821
edited Oct 4 at 8:32
pointguard0
1,248821
edited Oct 4 at 8:32
pointguard0
1,248821
1,248821
asked Oct 4 at 4:38
the-realtom
1185
asked Oct 4 at 4:38
the-realtom
1185
asked Oct 4 at 4:38
the-realtom
1185
1185
3
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
Oct 4 at 4:41
add a comment |Â
3
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
Oct 4 at 4:41
3
3
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
Oct 4 at 4:41
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
Oct 4 at 4:41
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
9
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 Oct 4 at 4:46
rpl19
2711
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
Oct 4 at 5:00
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
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 Oct 4 at 4:43
Ross Millikan
283k23191359
add a comment |Â
up vote
1
down vote
Another way to see this: suppose the 2 dice are thrown one after the other.
You're correct that your need either $(3,1)$ or $(1,3)$.
But the probability of each of these rolls is not $frac16timesfrac16$. In fact, since you can't have pairs, the first die can take 6 values, but the second one can only take five (all values, except the one of the first die).
So the probability of getting a "correct" roll is $$p(3,1)+p(1,3)=(frac16timesfrac15)+(frac16timesfrac15)=frac230 = frac115$$
answered Oct 4 at 12:59
F.Carette
9199
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
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 Oct 4 at 4:46
rpl19
2711
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
Oct 4 at 5:00
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
add a comment |Â
up vote
9
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 Oct 4 at 4:46
rpl19
2711
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
Oct 4 at 5:00
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
add a comment |Â
up vote
9
down vote
accepted
up vote
9
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 Oct 4 at 4:46
rpl19
2711
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 Oct 4 at 4:46
rpl19
2711
answered Oct 4 at 4:46
rpl19
2711
answered Oct 4 at 4:46
rpl19
2711
answered Oct 4 at 4:46
rpl19
2711
2711
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
Oct 4 at 5:00
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
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
Oct 4 at 5:00
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
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
Oct 4 at 5:00
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
Oct 4 at 5:00
3
3
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
It might be worth adding that the question is asking for the probability of A | B (A given B). Knowing that 'given' is a common keyword for these types of questions might help avoid similar mistakes in future.
– Cyberspark
Oct 4 at 12:26
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 Oct 4 at 4:43
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 Oct 4 at 4:43
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 Oct 4 at 4:43
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 Oct 4 at 4:43
Ross Millikan
283k23191359
answered Oct 4 at 4:43
Ross Millikan
283k23191359
answered Oct 4 at 4:43
Ross Millikan
283k23191359
answered Oct 4 at 4:43
Ross Millikan
283k23191359
283k23191359
add a comment |Â
add a comment |Â
up vote
1
down vote
Another way to see this: suppose the 2 dice are thrown one after the other.
You're correct that your need either $(3,1)$ or $(1,3)$.
But the probability of each of these rolls is not $frac16timesfrac16$. In fact, since you can't have pairs, the first die can take 6 values, but the second one can only take five (all values, except the one of the first die).
So the probability of getting a "correct" roll is $$p(3,1)+p(1,3)=(frac16timesfrac15)+(frac16timesfrac15)=frac230 = frac115$$
answered Oct 4 at 12:59
F.Carette
9199
add a comment |Â
up vote
1
down vote
Another way to see this: suppose the 2 dice are thrown one after the other.
You're correct that your need either $(3,1)$ or $(1,3)$.
But the probability of each of these rolls is not $frac16timesfrac16$. In fact, since you can't have pairs, the first die can take 6 values, but the second one can only take five (all values, except the one of the first die).
So the probability of getting a "correct" roll is $$p(3,1)+p(1,3)=(frac16timesfrac15)+(frac16timesfrac15)=frac230 = frac115$$
answered Oct 4 at 12:59
F.Carette
9199
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another way to see this: suppose the 2 dice are thrown one after the other.
You're correct that your need either $(3,1)$ or $(1,3)$.
But the probability of each of these rolls is not $frac16timesfrac16$. In fact, since you can't have pairs, the first die can take 6 values, but the second one can only take five (all values, except the one of the first die).
So the probability of getting a "correct" roll is $$p(3,1)+p(1,3)=(frac16timesfrac15)+(frac16timesfrac15)=frac230 = frac115$$
answered Oct 4 at 12:59
F.Carette
9199
Another way to see this: suppose the 2 dice are thrown one after the other.
You're correct that your need either $(3,1)$ or $(1,3)$.
But the probability of each of these rolls is not $frac16timesfrac16$. In fact, since you can't have pairs, the first die can take 6 values, but the second one can only take five (all values, except the one of the first die).
So the probability of getting a "correct" roll is $$p(3,1)+p(1,3)=(frac16timesfrac15)+(frac16timesfrac15)=frac230 = frac115$$
answered Oct 4 at 12:59
F.Carette
9199
answered Oct 4 at 12:59
F.Carette
9199
answered Oct 4 at 12:59
F.Carette
9199
answered Oct 4 at 12:59
F.Carette
9199
9199
add a comment |Â
add a comment |Â
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3
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
Oct 4 at 4:41