Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
$(i) \, 6$
$(ii) \, 12$
$(iii) \, 7$

(N/A) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
$(i)$ For the product to be $6$,the possible outcomes are $(1, 6), (2, 3), (3, 2), (6, 1)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(ii)$ For the product to be $12$,the possible outcomes are $(2, 6), (3, 4), (4, 3), (6, 2)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(iii)$ For the product to be $7$,there are no possible outcomes since $7$ is a prime number and cannot be formed by the product of two numbers from $1$ to $6$ (except $1 \times 7$,but $7$ is not on a die).
Number of favorable outcomes $= 0$.
Probability $= \frac{0}{36} = 0$.