Rolling a dice probability [closed]












-1












$begingroup$


What is the probability of getting



a) 3 distinct numbers (no order in the outcome e.g. 1,2,3 or 4,1,3 etc)



b) 2 distinct from rolling a die 6 times






probability permutations combinations







share|cite|improve this question











$endgroup$



closed as off-topic by lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber♦ Jan 17 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    This is very similar to your prior question. The solution you were given to that one should help you here as well.
    $endgroup$
    – lulu
    Jan 16 at 21:34










  • $begingroup$
    Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
    $endgroup$
    – jordan_glen
    Jan 16 at 21:37
















-1












$begingroup$


What is the probability of getting



a) 3 distinct numbers (no order in the outcome e.g. 1,2,3 or 4,1,3 etc)



b) 2 distinct from rolling a die 6 times






probability permutations combinations







share|cite|improve this question











$endgroup$



closed as off-topic by lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber♦ Jan 17 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    This is very similar to your prior question. The solution you were given to that one should help you here as well.
    $endgroup$
    – lulu
    Jan 16 at 21:34










  • $begingroup$
    Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
    $endgroup$
    – jordan_glen
    Jan 16 at 21:37














-1












-1








-1





$begingroup$


What is the probability of getting



a) 3 distinct numbers (no order in the outcome e.g. 1,2,3 or 4,1,3 etc)



b) 2 distinct from rolling a die 6 times






probability permutations combinations







share|cite|improve this question











$endgroup$




What is the probability of getting



a) 3 distinct numbers (no order in the outcome e.g. 1,2,3 or 4,1,3 etc)



b) 2 distinct from rolling a die 6 times






probability permutations combinations




probability permutations combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 21:36







nuune

















asked Jan 16 at 21:33









nuune nuune

73




73




closed as off-topic by lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber♦ Jan 17 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – lulu, David G. Stork, Arnaud D., 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 lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber♦ Jan 17 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – lulu, David G. Stork, Arnaud D., Shailesh, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    This is very similar to your prior question. The solution you were given to that one should help you here as well.
    $endgroup$
    – lulu
    Jan 16 at 21:34










  • $begingroup$
    Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
    $endgroup$
    – jordan_glen
    Jan 16 at 21:37














  • 2




    $begingroup$
    This is very similar to your prior question. The solution you were given to that one should help you here as well.
    $endgroup$
    – lulu
    Jan 16 at 21:34










  • $begingroup$
    Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
    $endgroup$
    – jordan_glen
    Jan 16 at 21:37








2




2




$begingroup$
This is very similar to your prior question. The solution you were given to that one should help you here as well.
$endgroup$
– lulu
Jan 16 at 21:34




$begingroup$
This is very similar to your prior question. The solution you were given to that one should help you here as well.
$endgroup$
– lulu
Jan 16 at 21:34












$begingroup$
Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
$endgroup$
– jordan_glen
Jan 16 at 21:37




$begingroup$
Possible duplicate of Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
$endgroup$
– jordan_glen
Jan 16 at 21:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

Some hints:



a) The first roll can be anything. What is the probability that the second roll is something different? What is the probability that the third roll is something different from the first two?



b) The first roll can be anything. The second roll can also be anything. If they're different, then the remaining four rolls must be one number or the other. If the first two numbers are the same, and the third number is different, then the remaining three rolls must be one number or the other. If the first three numbers are the same, and the fourth is different, then the remaining two rolls must be one number or the other. If the first four numbers are the same, and the fifth is different, then the remaining roll must be one number or the other. If the first five rolls are the same, the sixth must be different.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Some hints:



    a) The first roll can be anything. What is the probability that the second roll is something different? What is the probability that the third roll is something different from the first two?



    b) The first roll can be anything. The second roll can also be anything. If they're different, then the remaining four rolls must be one number or the other. If the first two numbers are the same, and the third number is different, then the remaining three rolls must be one number or the other. If the first three numbers are the same, and the fourth is different, then the remaining two rolls must be one number or the other. If the first four numbers are the same, and the fifth is different, then the remaining roll must be one number or the other. If the first five rolls are the same, the sixth must be different.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Some hints:



      a) The first roll can be anything. What is the probability that the second roll is something different? What is the probability that the third roll is something different from the first two?



      b) The first roll can be anything. The second roll can also be anything. If they're different, then the remaining four rolls must be one number or the other. If the first two numbers are the same, and the third number is different, then the remaining three rolls must be one number or the other. If the first three numbers are the same, and the fourth is different, then the remaining two rolls must be one number or the other. If the first four numbers are the same, and the fifth is different, then the remaining roll must be one number or the other. If the first five rolls are the same, the sixth must be different.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Some hints:



        a) The first roll can be anything. What is the probability that the second roll is something different? What is the probability that the third roll is something different from the first two?



        b) The first roll can be anything. The second roll can also be anything. If they're different, then the remaining four rolls must be one number or the other. If the first two numbers are the same, and the third number is different, then the remaining three rolls must be one number or the other. If the first three numbers are the same, and the fourth is different, then the remaining two rolls must be one number or the other. If the first four numbers are the same, and the fifth is different, then the remaining roll must be one number or the other. If the first five rolls are the same, the sixth must be different.






        share|cite|improve this answer









        $endgroup$



        Some hints:



        a) The first roll can be anything. What is the probability that the second roll is something different? What is the probability that the third roll is something different from the first two?



        b) The first roll can be anything. The second roll can also be anything. If they're different, then the remaining four rolls must be one number or the other. If the first two numbers are the same, and the third number is different, then the remaining three rolls must be one number or the other. If the first three numbers are the same, and the fourth is different, then the remaining two rolls must be one number or the other. If the first four numbers are the same, and the fifth is different, then the remaining roll must be one number or the other. If the first five rolls are the same, the sixth must be different.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 22:05









        JohnJohn

        22.8k32550




        22.8k32550















            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            Npm cannot find a required file even through it is in the searched directory

            Image
            up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more