Its Almost...

sum of any two odds is even everytime

any odd no. plus any other odd no. equals an even no.

The numbers on the dice are 1, 3, 5, 7, 9, 11, which are the first 6 consecutive numbers. Since sum of two odd numbers is always even, for two throws, the sum of outcomes will always be even and so the probability is 1.

Sum of two odd numbers is always an Even number, the dice contains only odd numbers so the sum is always even

represent an odd number as 2n+1 (2n+1)+(2n+1)=4n+2 which is always even

The sum of 2 odd numbers will always be equal to 1. Since each surface of the dice has an odd number, the sum of any 2 faces would be 1. Therefore the probability that the sum is even would be 1.

the ans is 1/2

every number will be odd, nad sum two odds is always an even number. Therefore 1

The sum of even is always even and sum of odds is also even we have 3 odd and 3 even on one dice. But we have two dices => we have 6 even and 6 odd numbers.

Thus P(e) = 12/12 = 1.