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Probability doubt - where am i wrong?

by jd12345
Tags: doubt, probability
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jd12345
#1
Apr27-12, 07:43 AM
P: 260
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?
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tiny-tim
#2
Apr27-12, 05:31 PM
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hi jd12345!
Quote Quote by jd12345 View Post
… but why cant we do this by favourable cases / total number of cases?
that formula only applies when each case is equally likely
ssd
#3
Apr30-12, 01:42 PM
P: 239
Check the three drawbacks of "Classical" definition.


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