Sum of a fair die

With two rolls, we have four possible outcomes:

Odd-Odd
Even-Even
Odd-Even
Even-Odd

Each of these has a probability of 1/4:

Odd-Odd = 1/2 * 1/2 = 1/4
Even-Even = 1/2 * 1/2 = 1/4
Odd-Even = 1/2 * 1/2 = 1/4
Even-Odd = 1/2 * 1/2 = 1/4

Which of them give us an even sum? Odd-Odd (1 + 3 = 4) and Even-Even ( 2 + 4 = 6) do, but Even-Odd and Odd-Even (1 + 2 = 3) do not. Thus, the probability of an even sum should be 1/4 + 1/4 = 1/2, which is not an answer choice. Is it a six-sided die numbered 1-6?

My mistake. The question is incomplete.

I think Anurag has it right.

@Anurag, is there a way to do this without listing everything out like that? i.e., using permutations + probability? Thanks.

My method didn't require writing anything out, just figuring out the possible arrangements and the probabilities for a given roll.

The simplest method of all:

In order to get a sum of even, you should either have even on first dice & even on second dice OR odd on first dice & odd on second dice.

P(Sum is even) = P(even on 1st)*P(even on 2nd) + P(odd on 1st)*P(odd on 2nd)
= (3/6)*(3/6) + (3/6)*(3/6)
= 1/2*1/2 + 1/2*1/2
= 1/4 + 1/4
= 1/2

Or, even intuitively, if you have equal number of even and odd numbers on a Dice, the Sum should be even in exactly half of the cases. So, the probability should be 50% or 1/2.

The intuition can be explained by the knowledge of
Sum being even in a (even X even) or (odd X odd) case,
and
Sum being odd in a (odd X even) or (even X odd) case.

So, out of four equally possible cases, 2 favour us. Therefore, Probability = 1/2.

Anyway, if this doesn't make too much sense, follow the method given above.

The sum of numbers from 1 to 6 taken 2 at a time is either even or odd, probability s 1/2 which is not in the option. Something is wrong with the question.