Mixing
When 2 balanced dice are rolled, what is the probability of the given numbers?
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When two balanced dice are rolled, what is the probability of getting:
- A) a sum of equal to 7,
- B) a sum of exactly equal to 11,
- C) a sum equal to zero?
probability dice
edited Feb 10 at 10:41
jvdhooft
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
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add a comment |
$begingroup$
When two balanced dice are rolled, what is the probability of getting:
- A) a sum of equal to 7,
- B) a sum of exactly equal to 11,
- C) a sum equal to zero?
probability dice
edited Feb 10 at 10:41
jvdhooft
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
$endgroup$
$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
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Add your own efforts on this to avoid downvotes and/or closing.
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– drhab
Feb 1 at 12:18
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I think in a case of two dice, you can just list all the possible outcomes.
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– Matti P.
Feb 1 at 12:20
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As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41
add a comment |
$begingroup$
When two balanced dice are rolled, what is the probability of getting:
- A) a sum of equal to 7,
- B) a sum of exactly equal to 11,
- C) a sum equal to zero?
probability dice
edited Feb 10 at 10:41
jvdhooft
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
$endgroup$
When two balanced dice are rolled, what is the probability of getting:
- A) a sum of equal to 7,
- B) a sum of exactly equal to 11,
- C) a sum equal to zero?
probability dice
probability dice
edited Feb 10 at 10:41
jvdhooft
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
edited Feb 10 at 10:41
jvdhooft
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
edited Feb 10 at 10:41
jvdhooft
5,65961641
edited Feb 10 at 10:41
jvdhooft
5,65961641
edited Feb 10 at 10:41
jvdhooft
5,65961641
5,65961641
asked Feb 1 at 12:14
Cess GoCess Go
1
asked Feb 1 at 12:14
Cess GoCess Go
1
asked Feb 1 at 12:14
Cess GoCess Go
1
1
$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
$begingroup$
Add your own efforts on this to avoid downvotes and/or closing.
$endgroup$
– drhab
Feb 1 at 12:18
$begingroup$
I think in a case of two dice, you can just list all the possible outcomes.
$endgroup$
– Matti P.
Feb 1 at 12:20
$begingroup$
As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41
add a comment |
$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
$begingroup$
Add your own efforts on this to avoid downvotes and/or closing.
$endgroup$
– drhab
Feb 1 at 12:18
$begingroup$
I think in a case of two dice, you can just list all the possible outcomes.
$endgroup$
– Matti P.
Feb 1 at 12:20
$begingroup$
As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41
$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
$begingroup$
Add your own efforts on this to avoid downvotes and/or closing.
$endgroup$
– drhab
Feb 1 at 12:18
$begingroup$
Add your own efforts on this to avoid downvotes and/or closing.
$endgroup$
– drhab
Feb 1 at 12:18
$begingroup$
I think in a case of two dice, you can just list all the possible outcomes.
$endgroup$
– Matti P.
Feb 1 at 12:20
$begingroup$
I think in a case of two dice, you can just list all the possible outcomes.
$endgroup$
– Matti P.
Feb 1 at 12:20
$begingroup$
As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41
$begingroup$
As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41
add a comment |
2 Answers
2
active
oldest
votes
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Total cases are $36$.
1. Favourable cases are $(1,6),(2,5),(3,4),(4,3),(5,2) & (6,1)$.
So, required probability $=frac{6}{36}=frac{1}{6}$.
2. Favourable cases are $(5,6) & (6,5)$.
So, required probability $=frac{2}{36}=frac{1}{18}$.
3. Required probability is $0$, as the minimum sum you can get is $2$, when both die show $1$.
Hope it is helpful:)
answered Feb 1 at 12:20
MartundMartund
1,965213
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add a comment |
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Assuming you mean standard 6 sided dice with the numbers 1 to 6 on each side, the possible pairs are (1, 1) which add to 2, (1,2) and (2, 1) which add to 3, (1, 3), (2, 2), and (3, 1) which add to 4, (1, 4), (2, 3), (3, 2), and (4, 1) which add to 5, (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1) which add to 6, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) which add to 7, (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2) which add to 8, (3, 6), (4, 5), (5, 4), and (6, 3) which add to 9, (4, 6), (5, 5), and (6, 4) which add to 10, (5, 6) and (6, 5) which add to 11, and, finally, (6, 6) which add to 12.
Of those 36 equally likely pairs, two, (5, 6) and (6, 5), add to 11 so the probability of 11 is 2/36= 1/18.
Of those 36 equally likely pairs, two, (1, 2) and (2, 1), add to 3 so the probability of 3 is 2/36= 1/18.
Of those 36 equally likely pairs, none add to a negative number so the probability of a negative number is 0/36= 0.
answered Feb 1 at 12:31
user247327user247327
11.6k1516
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
Total cases are $36$.
1. Favourable cases are $(1,6),(2,5),(3,4),(4,3),(5,2) & (6,1)$.
So, required probability $=frac{6}{36}=frac{1}{6}$.
2. Favourable cases are $(5,6) & (6,5)$.
So, required probability $=frac{2}{36}=frac{1}{18}$.
3. Required probability is $0$, as the minimum sum you can get is $2$, when both die show $1$.
Hope it is helpful:)
answered Feb 1 at 12:20
MartundMartund
1,965213
$endgroup$
add a comment |
$begingroup$
Total cases are $36$.
1. Favourable cases are $(1,6),(2,5),(3,4),(4,3),(5,2) & (6,1)$.
So, required probability $=frac{6}{36}=frac{1}{6}$.
2. Favourable cases are $(5,6) & (6,5)$.
So, required probability $=frac{2}{36}=frac{1}{18}$.
3. Required probability is $0$, as the minimum sum you can get is $2$, when both die show $1$.
Hope it is helpful:)
answered Feb 1 at 12:20
MartundMartund
1,965213
$endgroup$
add a comment |
$begingroup$
Total cases are $36$.
1. Favourable cases are $(1,6),(2,5),(3,4),(4,3),(5,2) & (6,1)$.
So, required probability $=frac{6}{36}=frac{1}{6}$.
2. Favourable cases are $(5,6) & (6,5)$.
So, required probability $=frac{2}{36}=frac{1}{18}$.
3. Required probability is $0$, as the minimum sum you can get is $2$, when both die show $1$.
Hope it is helpful:)
answered Feb 1 at 12:20
MartundMartund
1,965213
$endgroup$
Total cases are $36$.
1. Favourable cases are $(1,6),(2,5),(3,4),(4,3),(5,2) & (6,1)$.
So, required probability $=frac{6}{36}=frac{1}{6}$.
2. Favourable cases are $(5,6) & (6,5)$.
So, required probability $=frac{2}{36}=frac{1}{18}$.
3. Required probability is $0$, as the minimum sum you can get is $2$, when both die show $1$.
Hope it is helpful:)
answered Feb 1 at 12:20
MartundMartund
1,965213
answered Feb 1 at 12:20
MartundMartund
1,965213
answered Feb 1 at 12:20
MartundMartund
1,965213
answered Feb 1 at 12:20
MartundMartund
1,965213
1,965213
add a comment |
add a comment |
$begingroup$
Assuming you mean standard 6 sided dice with the numbers 1 to 6 on each side, the possible pairs are (1, 1) which add to 2, (1,2) and (2, 1) which add to 3, (1, 3), (2, 2), and (3, 1) which add to 4, (1, 4), (2, 3), (3, 2), and (4, 1) which add to 5, (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1) which add to 6, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) which add to 7, (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2) which add to 8, (3, 6), (4, 5), (5, 4), and (6, 3) which add to 9, (4, 6), (5, 5), and (6, 4) which add to 10, (5, 6) and (6, 5) which add to 11, and, finally, (6, 6) which add to 12.
Of those 36 equally likely pairs, two, (5, 6) and (6, 5), add to 11 so the probability of 11 is 2/36= 1/18.
Of those 36 equally likely pairs, two, (1, 2) and (2, 1), add to 3 so the probability of 3 is 2/36= 1/18.
Of those 36 equally likely pairs, none add to a negative number so the probability of a negative number is 0/36= 0.
answered Feb 1 at 12:31
user247327user247327
11.6k1516
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add a comment |
$begingroup$
Assuming you mean standard 6 sided dice with the numbers 1 to 6 on each side, the possible pairs are (1, 1) which add to 2, (1,2) and (2, 1) which add to 3, (1, 3), (2, 2), and (3, 1) which add to 4, (1, 4), (2, 3), (3, 2), and (4, 1) which add to 5, (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1) which add to 6, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) which add to 7, (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2) which add to 8, (3, 6), (4, 5), (5, 4), and (6, 3) which add to 9, (4, 6), (5, 5), and (6, 4) which add to 10, (5, 6) and (6, 5) which add to 11, and, finally, (6, 6) which add to 12.
Of those 36 equally likely pairs, two, (5, 6) and (6, 5), add to 11 so the probability of 11 is 2/36= 1/18.
Of those 36 equally likely pairs, two, (1, 2) and (2, 1), add to 3 so the probability of 3 is 2/36= 1/18.
Of those 36 equally likely pairs, none add to a negative number so the probability of a negative number is 0/36= 0.
answered Feb 1 at 12:31
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
Assuming you mean standard 6 sided dice with the numbers 1 to 6 on each side, the possible pairs are (1, 1) which add to 2, (1,2) and (2, 1) which add to 3, (1, 3), (2, 2), and (3, 1) which add to 4, (1, 4), (2, 3), (3, 2), and (4, 1) which add to 5, (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1) which add to 6, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) which add to 7, (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2) which add to 8, (3, 6), (4, 5), (5, 4), and (6, 3) which add to 9, (4, 6), (5, 5), and (6, 4) which add to 10, (5, 6) and (6, 5) which add to 11, and, finally, (6, 6) which add to 12.
Of those 36 equally likely pairs, two, (5, 6) and (6, 5), add to 11 so the probability of 11 is 2/36= 1/18.
Of those 36 equally likely pairs, two, (1, 2) and (2, 1), add to 3 so the probability of 3 is 2/36= 1/18.
Of those 36 equally likely pairs, none add to a negative number so the probability of a negative number is 0/36= 0.
answered Feb 1 at 12:31
user247327user247327
11.6k1516
$endgroup$
Assuming you mean standard 6 sided dice with the numbers 1 to 6 on each side, the possible pairs are (1, 1) which add to 2, (1,2) and (2, 1) which add to 3, (1, 3), (2, 2), and (3, 1) which add to 4, (1, 4), (2, 3), (3, 2), and (4, 1) which add to 5, (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1) which add to 6, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) which add to 7, (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2) which add to 8, (3, 6), (4, 5), (5, 4), and (6, 3) which add to 9, (4, 6), (5, 5), and (6, 4) which add to 10, (5, 6) and (6, 5) which add to 11, and, finally, (6, 6) which add to 12.
Of those 36 equally likely pairs, two, (5, 6) and (6, 5), add to 11 so the probability of 11 is 2/36= 1/18.
Of those 36 equally likely pairs, two, (1, 2) and (2, 1), add to 3 so the probability of 3 is 2/36= 1/18.
Of those 36 equally likely pairs, none add to a negative number so the probability of a negative number is 0/36= 0.
answered Feb 1 at 12:31
user247327user247327
11.6k1516
answered Feb 1 at 12:31
user247327user247327
11.6k1516
answered Feb 1 at 12:31
user247327user247327
11.6k1516
answered Feb 1 at 12:31
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
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$begingroup$
What are the numbers on your dice? I am asking because you want the sum to be $0$...
$endgroup$
– Klaus
Feb 1 at 12:16
$begingroup$
Add your own efforts on this to avoid downvotes and/or closing.
$endgroup$
– drhab
Feb 1 at 12:18
$begingroup$
I think in a case of two dice, you can just list all the possible outcomes.
$endgroup$
– Matti P.
Feb 1 at 12:20
$begingroup$
As asked by Klaus, are they standard dice? And why do you differentiate betwen "equal to 7" and " exactly equal to 11"
$endgroup$
– Thomas Lesgourgues
Feb 1 at 12:41