The odds in favour of getting sum multiple of 3, when pair of dice are thrown is
The odds in favour of getting sum multiple of 3, when pair of dice are thrown is
$4: 5$
$2: 3$
$1: 2$
$3: 4$
Solution
When a pair of dice is thrown, then total outcomes are $6 \times 6=36$.
Now the odds in favour of getting the sum, which is multiple of 3 are
$(1,2)(2,1) \Rightarrow \operatorname{sum} 3$
$(3,3),(2,4),(4,2),(1,5),(5,1) \Rightarrow \operatorname{sum} 6$
$(4,5),(5,4),(6,3),(3,6) \Rightarrow \operatorname{sum} 9$
$(6,6) \Rightarrow$ sum 12
Thus Number of favourable cases $=2+5+4+1=12$
So, odds in fovour $=\frac{12}{24}=\frac{1}{2}$