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An octahedral die whose faces are numbered 1 through 8 (only one number on one face) is thrown three times. What is the probability that the product of the numbers obtained in first two throws is equal to the number obtained in the third throw? (1) 9 / 216 (2) 3 / 128 (3) 3 / 64 (4) 5 / 128 |
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Answer» (4) 5 / 128 Total possible outcomes = 8 × 8 × 8 = 512 Favourable outcomes = {(1, 1, 1), (1, 2, 2), (2, 1, 2), (1, 3, 3), (3, 1, 3), (1, 4, 4), (4, 1, 4), (2, 2, 4), (1, 5, 5), (5, 1, 5), (1, 6, 6), (6, 1, 6), (2, 3, 6), (3, 2, 6), (1, 7, 7), (7, 1, 7), (8, 1, 8), (1, 8, 8), (2, 4, 8), (4, 2, 8) } = 20 Required Probability = 20 / 512 = 5 / 128 |
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