Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

(i) 6

(ii) 12

(iii) 7

Number of total outcomes = 36

(i) When product of the numbers on the top of the dice is 6.

So, the possible ways are (1, 6), (2,3), (3, 2) and (6, 1).

Number of possible ways = 4

∴ Required probability = 4/36 = 1/9

(ii) When product of the numbers on the top of the dice is 12.

So, the possible ways are (2, 6), (3,4), (4, 3) and (6, 2).

Number of possible ways = 4

∴ Required probability = 4/36 = 1/9

(iii) Product of the numbers on the top of the dice cannot be 7. So, its probability is zero.