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.


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