Question

Two dice are thrown. Find the odds in favour of getting the sum $4$.

• A. $1:11$
• B. $11:1$
• C. $3:11$
• D. $11:3$

Answer 1

Answer: (A)

$1:11$

Total number of exhaustive cases=$6\times 6=36$
A sum of 4 can be obtained as $(1,3)(2,2)(3,1)$
Therefore, there are 3 favorable outcomes and $(36-3)=33$ unfavorable outcomes.
$\therefore$ Odds in favour of sum of 4 $=\frac{3}{33}=\frac{1}{11}$.