Two balanced dice are thrown simultaneously. So the sample space for this random experiment Is expressed as follows:

U ((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)}

∴ Total number of primary outcomes of U is
n = 36

(1) A = Event that the sum of numbers on the dice is 6.
A = {( 1, 5), (2. 4), (3, 3), (4, 2). (5. 1))

∴ Favourable outcomes for the event A is m = 5.

Hence, P(A) = \(\frac{m}{n} = \frac{5}{36}\)

(2) B = Event that the sum of numbers on the dice is more than 10.

∴ B’ = Event that the sum of numbers on the dice is not more than 10.

B = {(5, 6), (6, 5), (6, 6)}

∴ Favourable outcomes for the event B is m = 3.

Hence P(B) = \(\frac{m}{n} = \frac{3}{36} = \frac{1}{12}\)

∴ P(B’) = 1 – P(B)

= 1 – \(\frac{1}{12} = \frac{11}{12}\)

(3) C = Event that the sum of numbers on the dice is a multiple of 3, i.e., 3, 6, 9 or 12

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

∴ Favourable outcomes for the event C is m = 12

Hence, P(C) =\( \frac{m}{n} = \frac{12}{36} = \frac{1}{3}\)

(4) D = Event that the product of the numbers on the dice is 12
D = {(2, 6), (3, 4), (4, 3), (6, 2)}

∴ Favourable outcomes for the event D is m = 4

Hence, P(D) =\( \frac{m}{n} = \frac{4}{36} = \frac{1}{9}\)