Question

Two dice are thrown together. The probability of getting the sum of digits as a multiple of 4 is:

• A. $\frac{1}{9}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{5}{9}$

Answer 1

Answer: (C)

$\frac{1}{4}$

Answer 2

Total exhaustive cases $= 6^2 = 36$ Following 9 pairs are favourable as the sum of their digits are multiple of 4 i.e., 4 or 8 or 12 (1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6) $\therefore$ Required probability $= \frac{9}{36} = \frac{1}{4}$