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  • a tough little problem

    Anyone know how to solve this problem. The answer is 1554. Thanks.

    Let X be a random variable representing the number of times you need to roll a dice (6 sided) until you get 4 consecutive 6's. Find E(X).

    - Matt

  • #2
    Tough little problem

    You must not be counting the last roll. I am getting 1555. Take a look at:
    http://www.math.ilstu.edu/krzysio/6-...O-Exercise.pdf
    But if I am wrong and it is not 1555, when the last roll is included, I welcome
    everyone's corrections.

    Yours,
    Krzys' Ostaszewski


    Originally posted by engelbmj
    Anyone know how to solve this problem. The answer is 1554. Thanks.

    Let X be a random variable representing the number of times you need to roll a dice (6 sided) until you get 4 consecutive 6's. Find E(X).

    - Matt
    Want to know how to pass actuarial exams? Go to: smartURL.it/pass

    Comment


    • #3
      Can anyone else access this link? I cannot. I get "page not found". If you figure it out, please share. Thanks.

      Originally posted by krzysio
      You must not be counting the last roll. I am getting 1555. Take a look at:
      http://www.math.ilstu.edu/krzysio/6-...O-Exercise.pdf
      But if I am wrong and it is not 1555, when the last roll is included, I welcome
      everyone's corrections.

      Yours,
      Krzys' Ostaszewski

      Comment


      • #4
        No, I cannot get the link to work either. I sent him a message to let him know.

        I am not sure how you even started this problem. Can you show me your work?

        Thanks,
        Karen

        Comment


        • #5
          Link access

          There was a problem with the server, it should be up now, sorry. Please try again and let me know if it works.

          Yours,
          Krzys' Ostaszewski

          Originally posted by engelbmj
          Can anyone else access this link? I cannot. I get "page not found". If you figure it out, please share. Thanks.
          Want to know how to pass actuarial exams? Go to: smartURL.it/pass

          Comment

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