From these 36 possible rolls:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)  

we remove all the ones that have sum 2, 4, or 6 

      (1,2)       (1,4)       (1,6)

(2,1)       (2,3)       (2,5) (2,6) 

      (3,2)       (3,4) (3,5) (3,6) 

(4,1)       (4,3) (4,4) (4,5) (4,6) 

      (5,2) (5,3) (5,4) (5,5) (5,6) 

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

I count that there are 27 remaining that
do not have sum 2, 4, or 6.

The probability is 27 out of 36 or 27/36
which reduces to 3/4.

Edwin