? Two dice are thrown together. What is the probability that the number obtained on one of the dice is multiple of number obtained on the other dice

Total number of cases = 62 = 36
Since the number on a die should be multiple of the other, the possibilities are
(1, 1) (2, 2) (3, 3) ------ (6, 6) --- 6 ways
(2, 1) (1, 2) (1, 4) (4, 1) (1, 3) (3, 1) (1, 5) (5, 1) (6, 1) (1, 6) --- 10 ways
(2, 4) (4, 2) (2, 6) (6, 2) (3, 6) (6, 3) -- 6 ways
Favorable cases are = 6 + 10 + 6 = 22. So, P (A) = 22/36 = 11/18