Two dice are thrown simultaneously. What is the probability that the sum of numbers on the two faces is divisible by 4 or 6?

$\frac{7}{18}$ $\frac{5}{36}$ $\frac{11}{18}$ $\frac{1}{6}$

$\frac{7}{18}$

The correct answer is Option (1) → $\frac{7}{18}$ Total Sets = 36 Required Sets = {(1,3) (3,1) (2,2) (1,5) (5,1) (3,3) (2,4) (4,2) (2,6) (6,2) (4,4) (3,5) (5,3) (6,6)} = 14 Thus, Required probability = $\frac{14}{36}$ = $\frac{7}{18}$