(a) 1 : 11
Total number of exhaustive cases = 6 × 6 = 36 A sum of 4 can be obtained as (1, 3) (2, 2) (3, 1) Therefore, there are 3 favourable outcomes and (36 – 3) = 33 unfavourable outcomes.
\(\therefore\) Odds in favour of sum of 4= \(\frac{3}{33}\) = \(\frac{1}{11}\)
