

InterviewSolution
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In a single throw of two dice, find the probability of(i) getting a sum less than 6(ii) getting a doublet of odd numbers(iii) getting the sum as a prime number |
Answer» Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes) Possible outcomes are as follow: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Total number of outcomes = 36 (i) getting a sum less than 6 Pick entries having sum less than 6: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) Total number of favorable outcomes = 10 Probability (getting a sum less than 6) = 10/36 or 5/18 (ii) getting a doublet of odd numbers Pick entries having doublet of odd numbers: (1, 1), (3, 3), (5, 5) Total number of favorable outcomes = 3 Probability (getting a doublet of odd numbers) = 3/36 or 1/12 (iii) getting the sum as a prime number Pick entries having sum as a prime number: (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) Total number of favorable outcomes = 15 Probability (getting the sum as a prime number) = 15/36 or 5/12 |
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