Multiple choice

Six faces of a die are marked with numbers 1, -1, 0, -2, 2, 3 and the die is thrown thrice. The probability that the sum of the numbers thrown is six, is

  1. $\dfrac{3}{216}$
  2. $\dfrac{6}{216}$
  3. $\dfrac{10}{216}$
  4. $\dfrac{18}{216}$
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C Correct answer
Explanation

Total outcomes = 6^3 = 216. We need combinations of (x, y, z) from {1, -1, 0, -2, 2, 3} that sum to 6. Possible sets: (3, 3, 0) [3 permutations], (3, 2, 1) [6 permutations], (2, 2, 2) [1 permutation]. Total = 3 + 6 + 1 = 10. Probability = 10/216.

AI explanation

The total number of possible outcomes when the die is thrown thrice is 6 * 6 * 6 = 216. We must find the combinations of the numbers 1, -1, 0, -2, 2, 3 that add up to exactly 6. The valid unordered sets of numbers are (3, 3, 0) with 3 permutations, (3, 2, 1) with 6 permutations, (3, 3, 0) has 3 ways, and (2, 2, 2) has 1 way, totaling 10 favorable outcomes. The probability is therefore 10/216.

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