if two dices are thrown, the total possible outcomes will be:
(1,1) , (2,1) , (3,1) , (4,1) , (5,1) , (6,1)
(1,2) , (2,2) , (3,2) , (4,2) , (5,2) , (6,2)
(1,3) , (2,3) , (3,3) , (4,3) , (5,3) , (6,3)
(1,4) , (2,4) , (3,4) , (4,4) , (5,4) , (6,4)
(1,5) , (2,5) , (3, 5) , (4,5) , (5,5) , (6,5)
(1,6) , (2,6) , (3,6) , (4,6) , (5,6) , (6,6)
now, total possible outcomes are 36
(i) outcomes = (2+6) , (3+5), (4+4) , (5+3), (6+2)
P(SUM=8) = `5/36`
(ii) outcomes= 0
P(SUM=13) = `0/36 = 0`
(iii) outcomes= (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)
P(SUM<=12) = `36/36 = 1`