Why can't I use favorable cases for a biased die?

[jd12345]

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 can't 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 can't we do this by favourable cases / total number of cases?

 

[tiny-tim]

hi jd12345!

… but why can't we do this by favourable cases / total number of cases?

that formula only applies when each case is equally likely

 

[ssd]

Check the three drawbacks of "Classical" definition.