Question

What are the odds in favour of getting sum multiple of 3, when pair of dice is thrown?

Answer 1

Solution:

When a pair of dice is thrown than total outcomes are
$( {6)}^{2} = 36$
Now the odds in favour of getting the sum,which is multiple of 3 are

(1,2) (2,1) => sum 3

(3,3) (2,4) (4,2) (1,5) (5,1) => sum 6

(4,5) (5,4) (6,3) (3,6) => sum 9

(6,6) => sum 12

all the above discussed cases are in favour of sum on both dice which are multiple of 3

Total in number= 12

So, odds in favour 12:24

$Probability =\frac{12}{36} \\ \\ = \frac{1}{3}$
Hope it helps you.