two dice are thrown together. what is the probability that the numbers on the two faces is neither divisible by 3 nor by 4?

two dice are thrown together , the possible sum { 2, 3,4, 5, ............, 12}
the total outcomes = 6*6 = 36
the number which are divisible by 3 or 4 is { 3, 4, 6, 8,9, 12}
the outcomes when number is divisible by 3 or 4 is
{(1,2);(2,1);(1,3);(3,1);(2,2);(1,5);(5,1);(2,4);(4,2);(3,3); (2,6);(6,2);(3,5);(5,3);(4,4);
(3,6);(6,3);(4,5);(5,4);(6,6)}
total number of case = 20
therefore the probability that the number is divisible by 3 or 4 is 20/36 = 5/9
thus the probability that the number is neither divisible by 3 nor by 4
$1-\frac{5}{9}=\frac{4}{9}$
hope this helps you