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Two dice are thrown. Find the odds in favour of getting the sum 4.(a) 1 : 11 (b) 11 : 1 (c) 4 : 11 (d) 11: 4 |
Answer» (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= \(\frac{3}{33}\) = \(\frac{1}{11}\) |
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