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Its a very good question.This is what I think :-
If we follow probabilistic methods given in text books, then method 1 is correct
in first 9 natural numbers, 4 even and 5 odd numbers are there.
ways of selecting 2 digits from 9 digits is : 9C2
sum of 2 even numbers is always even, and sum of 2 odd numbers is always even
ways of slecting this is : 4C2 + 5C2
probability is : 4C2 + 5C2 / 9C2 = 4/9
--
I'm giving an other example to support this statement,
now to find probability of any 3 numbers selected br even from a set of 9 natural numbers is ...
ways of selecting 3 digits from 9 digits is : 9C3
sum of 3 even numbers is always even, and sum of 2 odd numbers,1 even number is always even
ways of slecting this is : 4C3 + 5C2*4C1
probability is : 4C3 + 5C2.4C1 / 9C3 = 44/84
It is obvious that here the probability as said by method 2 cannot be 1/2
Regards
Ramesh
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