3 (six-sided) dice are rolled. What is the probability that the sum of the numbers is less than or equal to...
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I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.
probability
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
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closed as off-topic by Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber♦ Dec 4 '18 at 3:37
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$begingroup$
I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.
probability
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
$endgroup$
closed as off-topic by Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber♦ Dec 4 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
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– EvanHehehe
Dec 4 '18 at 0:53
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$begingroup$
I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.
probability
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
$endgroup$
I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.
probability
probability
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
asked Dec 4 '18 at 0:49
Alex SpielmanAlex Spielman
1
1
closed as off-topic by Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber♦ Dec 4 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber♦ Dec 4 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Saad, Chase Ryan Taylor, Shailesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
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– EvanHehehe
Dec 4 '18 at 0:53
add a comment |
2
$begingroup$
You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:53
2
2
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You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
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– EvanHehehe
Dec 4 '18 at 0:53
$begingroup$
You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:53
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1 Answer
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The sum can be either 3, 4, or 5.
If the sum is 3, the dice must be $1, 1, 1$, with probability $displaystyle left(frac{1}{6}right)^3$
If the sum is 4, the dice must be $1, 1, 2$. This dice rolls can be $1,2,1$ or $2,1,1$ (ordering), so the probability is $displaystyle 3left(frac{1}{6}right)^3$
If the sum is 5, the dice must be $1, 1, 3$ or $1, 2, 2$, similarly both cases have probability $3displaystyle left(frac{1}{6}right)^3$.
So, the total probability is $displaystyle boxed{frac{10}{216}}$.
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sum can be either 3, 4, or 5.
If the sum is 3, the dice must be $1, 1, 1$, with probability $displaystyle left(frac{1}{6}right)^3$
If the sum is 4, the dice must be $1, 1, 2$. This dice rolls can be $1,2,1$ or $2,1,1$ (ordering), so the probability is $displaystyle 3left(frac{1}{6}right)^3$
If the sum is 5, the dice must be $1, 1, 3$ or $1, 2, 2$, similarly both cases have probability $3displaystyle left(frac{1}{6}right)^3$.
So, the total probability is $displaystyle boxed{frac{10}{216}}$.
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
$endgroup$
add a comment |
$begingroup$
The sum can be either 3, 4, or 5.
If the sum is 3, the dice must be $1, 1, 1$, with probability $displaystyle left(frac{1}{6}right)^3$
If the sum is 4, the dice must be $1, 1, 2$. This dice rolls can be $1,2,1$ or $2,1,1$ (ordering), so the probability is $displaystyle 3left(frac{1}{6}right)^3$
If the sum is 5, the dice must be $1, 1, 3$ or $1, 2, 2$, similarly both cases have probability $3displaystyle left(frac{1}{6}right)^3$.
So, the total probability is $displaystyle boxed{frac{10}{216}}$.
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
$endgroup$
add a comment |
$begingroup$
The sum can be either 3, 4, or 5.
If the sum is 3, the dice must be $1, 1, 1$, with probability $displaystyle left(frac{1}{6}right)^3$
If the sum is 4, the dice must be $1, 1, 2$. This dice rolls can be $1,2,1$ or $2,1,1$ (ordering), so the probability is $displaystyle 3left(frac{1}{6}right)^3$
If the sum is 5, the dice must be $1, 1, 3$ or $1, 2, 2$, similarly both cases have probability $3displaystyle left(frac{1}{6}right)^3$.
So, the total probability is $displaystyle boxed{frac{10}{216}}$.
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
$endgroup$
The sum can be either 3, 4, or 5.
If the sum is 3, the dice must be $1, 1, 1$, with probability $displaystyle left(frac{1}{6}right)^3$
If the sum is 4, the dice must be $1, 1, 2$. This dice rolls can be $1,2,1$ or $2,1,1$ (ordering), so the probability is $displaystyle 3left(frac{1}{6}right)^3$
If the sum is 5, the dice must be $1, 1, 3$ or $1, 2, 2$, similarly both cases have probability $3displaystyle left(frac{1}{6}right)^3$.
So, the total probability is $displaystyle boxed{frac{10}{216}}$.
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
answered Dec 4 '18 at 0:56
Saketh MalyalaSaketh Malyala
7,3451534
7,3451534
add a comment |
add a comment |
2
$begingroup$
You can just start by calculating the probability of the sum equal to 3, then 4, then 5. And add them all up.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:53