Two dice are rolled by two players A and B. A throws total 10, the probability that B throws more than A is ______.
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Two dice are rolled by two players A and B. A throws total 10, the probability that B throws more than A is ______.
1/12
1/36
1.18
None of the above
If A throws a 10, the possible outcomes for A are (4,6), (5,5), (6,4). B needs to throw more than 10 (i.e., 11 or 12). The outcomes for B to get 11 are (5,6), (6,5) and for 12 is (6,6). Total outcomes for two dice are 36. Probability = 3/36 = 1/12.
When two dice are rolled, the total number of outcomes is 36, and B must roll an 11 or 12 to throw more than A's total of 10. A sum of 11 can be achieved in two ways, (5, 6) and (6, 5), and a sum of 12 can be achieved in one way, (6, 6). Thus, there are 3 favorable outcomes out of 36, making the probability 3 divided by 36, which is 1/12.