Two fair and ordinary dice are rolled simultaneously. The probability of getting the sum of outcomes of the dice as a multiple of 4, is equal to

$\frac{1}{4}$

In the case sum can be 4 or 8 or 12.

Total ways of getting the sum as 4

= 3 (namely (2, 2), (1, 3), (3, 1)).

Total ways of getting the sum as 8

= 5 (namely (2, 6), (6, 2), (3, 5), (5, 3), (4, 4))

Total ways of getting the sum as 12

= 1 (namely (6, 6))

Thus total ways of getting the sum as multiple of four

= 3 + 5 + 1 = 9

Thus, required probability = $\frac{9}{36}=\frac{1}{4}$