if a die is rolled. what is the probability of getting sum greater or equal to 10
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For the 3 rolls to sum to 10, the sum of the 1st 2 rolls must be from 4 to 9. Let n(j) be the number of ways two dice can sum to j. Then n(4) = 3, since the first roll can be an integer from 1 to 3. By similar logic, n(5) = 4, n(6) = 5, n(7) = 6, n(8) = 5, and n(9) = 4. Since the sum of n(i) for i from 4 to 9 equals 27, the probability of 3 rolls summing to 10 is 27/(6**3) = 27/216 = 1/8.
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For the 3 rolls to sum to 10, the sum of the 1st 2 rolls must be from 4 to 9. Let n(j) be the number of ways two dice can sum to j. Then n(4) = 3, since the first roll can be an integer from 1 to 3. By similar logic, n(5) = 4, n(6) = 5, n(7) = 6, n(8) = 5, and n(9) = 4. Since the sum of n(i) for i from 4 to 9 equals 27, the probability of 3 rolls summing to 10 is 27/(6**3) = 27/216 = 1/8.
Thank you
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