hsharif2 Wrote:Since there are 150 evens and 150 odds, if you were to choose one number the chances that it would be even OR odd would be 50%
Wouldn't it make sense to conclude that the sum of three numbers would also have an equal probability of being even or odd? Probability wouldn't favor odd or even.
not necessarily.
if you could reason this way, then the reasoning would generalize to all symmetrical situations -- and it doesn't.
for instance, this reasoning would also lead to the prediction that there would be a 50% chance of "odd" for the sum of two numbers out of a set of four.
but, if you have four balls 1, 2, 3, 4 and you pick two of them, then four of the six possibilities (1 and 2, 1 and 4, 2 and 3, 3 and 4) give odd sums, and only two of them (1 and 3, 2 and 4) give even sums. so, in that case, your sum has a 2/3 chance of being odd and only 1/3 chance of being even.
more generally, the flaw in your reasoning is that you aren't acknowledging the non-replacement of the balls from the box.
i.e., once you have removed one of the balls, then the situation is no longer symmetrical, so there's no reason to expect the probabilities to turn out symmetrically.
