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Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8?(a) \(\frac{32}{87}\)(b) \(\frac{12}{43}\)(c) \(\frac{6}{13}\)(d) \(\frac{23}{64}\)This question was posed to me during an interview.Enquiry is from Discrete Probability topic in division Discrete Probability of Discrete Mathematics |
Answer» CORRECT choice is (d) \(\frac{23}{64}\) Easy explanation: Sample space CONSISTS of 8*8=64 events. While (8) has \(\frac{1}{8}\) probability of occurrence, (1,7) has only \(\frac{1}{64}\) probability. So, the required probability = \(\frac{1}{6} + (9 * \frac{1}{64}) = \frac{69}{192} = \frac{23}{64}\). |
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