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

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

Probability

Solution:

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}$