If A represent a set of first 9 natural numbers then find probability that sum of any two numbers from set A is even.Method 1:(4C2+5C2)/9C2=4/9method 2:the probability should be half because the sum is either even or odd.What's the fallacy.

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

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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

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Regards

Ramesh