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

Question

Two dice are thrown together. The probability of getting the sum of digits as a multiple of 4 is: (A) (1/9) (B) (1/3) (C) (1/4) (D) (5/9)

Tardigrade Admin · Accepted Answer

Total exhaustive cases = 62 = 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) ∴ Required probability = (9/36) = (1/4)

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

$\frac{1}{9}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{5}{9}$

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)
$∴$ Required probability $= \frac{9}{36} = \frac{1}{4}$