Two dice are thrown together.

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 $=4$

$\therefore \quad$ Required probability $=\frac{4}{36}=\frac{1}{9}$

(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 $=4$

$\therefore \quad$ Required probability $=\frac{4}{36}=\frac{1}{9}$

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