Probability doubt - where am i wrong?

  1. There is a dice where probability of obtaining a number is proportional to the number itself.
    I have done these problems and the probabilties are 1/21, 2/21,/3/21,/4/21,/5/21 and 6/21

    In a normal dice when you have to find probabilities of say, 2 and 4 we do it by favourable cases / total number of cases = 2/6 = 1/3

    But you cant apply that method in this case - i know that but why?

    Probability of 2 or 4 in this case would be 2/21 + 4/21 = 6/21
    but why cant we do this by favourable cases / total number of cases?
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    hi jd12345! :smile:
    that formula only applies when each case is equally likely :wink:
     
  4. Check the three drawbacks of "Classical" definition.
     
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