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In a single throw of two dice, find the probability of getting a doublet of odd numbers.(a) \(\frac{1}{9}\)(b) \(\frac{1}{18}\)(c) \(\frac{1}{36}\)(d) \(\frac{1}{12}\) |
Answer» (d) \(\frac{1}{12}\) Total number of exhaustive cases in a single throw of two dice = 6 × 6 = 36 Doublets are obtained as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) Number of doublets of odd numbers = 3 \(\therefore\) Required probability = \(\frac{3}{36}\) = \(\frac{1}{12}\) |
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