Let ‘S’ be the sample space. 

∴ 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)}. 

⇒ n(S) = 36 

Let ‘A’ and ‘B’ be the events that the sum of the numbers on the two faces is divisible by 3 and 4 respectively. 

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

⇒ n(A) = 12 

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

⇒ n(B) = 9

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) = \(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) = \(\frac{n(B)}{n(S)}=\frac{9}{36}\)

A ∩ B = {(6, 6)} ⇒ n(A ∩ B) = 1

∴ (A ∩ B) = \(\frac{n(A∩ B)}{n(S)}=\frac{1}{36}\)

We know– 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{13}{36}+\frac{9}{36}-\frac{1}{36}=\frac{20}{36}=\frac{5}{9}\)

∴ Probability of getting the sum multiple of 3 or 4 is \(\frac{5}{9}\)

⇒ Probability of getting the sum neither a multiple of 3 nor 4 = P(A ∪ B)

= \(p(\overline{A∪B})\)

= 1 – P(A ∪ B)

\(= 1 - \frac{5}{9}\)

= \(\frac{4}{9}\)