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. Solution to Problem 1.14. (a) Each possible outcome has probability 1/36. There are 6 possible outcomes that are doubles, so the probability of doubles is 6=36 = 1=6. (b) The conditioning event (sum is 4 or less) consists of the 6 outcomes (1;1);(1;2);(1;3);(2;1);(2;2);(3;1); 2 of which are doubles, so the conditional probability of doubles is 2=6 = 1=3., Solution to Problem 1.14. (a) Each possible outcome has probability 1/36. There are 6 possible outcomes that are doubles, so the probability of doubles is 6=36 = 1=6. (b) The conditioning event (sum is 4 or less) consists of the 6 outcomes (1;1);(1;2);(1;3);(2;1);(2;2);(3;1); 2 of which are doubles, so the conditional probability of doubles is 2=6 = 1=3..

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. Introduction to Probability Models, Student Solutions Manual (e-only): Introduction to Probability Models 10th Edition Sheldon M Ross, Solution to Problem 1.14. (a) Each possible outcome has probability 1/36. There are 6 possible outcomes that are doubles, so the probability of doubles is 6=36 = 1=6. (b) The conditioning event (sum is 4 or less) consists of the 6 outcomes (1;1);(1;2);(1;3);(2;1);(2;2);(3;1); 2 of which are doubles, so the conditional probability of doubles is 2=6 = 1=3..

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