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

(i) 4

(ii) 12

(iii) 7

When two different dice are rolled together, the total number of outcomes is 6 × 6 i.e. 36 and all outcomes are equally likely. The sample space of the random experiment has 36 equally likely outcomes. The sample of the experiment

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
       (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
       (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
       (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
       (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).}

It consists of 36 equally likely outcomes.

(i) Let A be the event of getting numbers on top of dice with product 4.

A = {(1, 4), (2, 2), (4, 1)}.

∴ P(A) = $\dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability that the product of the numbers on the top of two dice is 4 is $\dfrac{1}{12}.$

(ii) Let B be the event of getting numbers on top of dice with product 12.

B = {(2, 6), (3, 4), (4, 3), (6, 2)}.

∴ P(B) = $\dfrac{4}{36} = \dfrac{1}{9}.$

Hence, the probability that the product of the numbers on the top of two dice is 12 is $\dfrac{1}{9}.$

(iii) Let C be the event of getting numbers on top of dice with product 7.

C = {}.

∴ P(C) = $\dfrac{0}{36} = 0.$

Hence, the probability that the product of the numbers on the top of two dice is 7 is 0.