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.






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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:


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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
















0












$begingroup$


I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.






probability







share|cite|improve this question









$endgroup$



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




    $begingroup$
    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














0












0








0





$begingroup$


I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.






probability







share|cite|improve this question









$endgroup$




I understand simpler probability problems, but got tripped up with this one. Please explain your work if possible. Thanks.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




    $begingroup$
    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




    $begingroup$
    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




$begingroup$
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




$begingroup$
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










1 Answer
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$begingroup$

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 $displaystyle boxed{frac{10}{216}}$.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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 $displaystyle boxed{frac{10}{216}}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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 $displaystyle boxed{frac{10}{216}}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $displaystyle boxed{frac{10}{216}}$.






        share|cite|improve this answer









        $endgroup$



        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 $displaystyle boxed{frac{10}{216}}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 0:56









        Saketh MalyalaSaketh Malyala

        7,3451534




        7,3451534















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