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If three dice are thrown simultaneously, find the probability of getting a sum ofat least 5 |
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Answer» Three dice thrown total events would be 6 X 6 X 6 = 216 Number of events of getting a total of 5 = 6 i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2) Therefore, PROBABILITY of getting a total of 5 Number of FAVORABLE outcomes P(E1) = Total number of possible outcome = 6/216 = 1/36 |
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