Multiple choice

On a symmetrical die the numbers 1, -1, 2, -2, 3 and 0 are marked on its 6 faces. If such a die is thrown 3 times, the probability that the sum of points on them is 6 is

  1. $\dfrac{5}{27}$
  2. $\dfrac{5}{54}$
  3. $\dfrac{5}{108}$
  4. $\dfrac{3}{108}$
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C Correct answer
Explanation

Total outcomes = 6^3 = 216. We need the sum of 3 faces to be 6. Possible combinations (order matters): (3,3,0)x3, (3,2,1)x6, (3,2,-1)x6, (2,2,2)x1, (2,2,0)x3, (2,1,1)x3, (3,1,2) is covered. Summing valid permutations leads to 10/216 = 5/108.

AI explanation

The total number of possible outcomes when the die is thrown 3 times is 6 cubed, or 216. To obtain a sum of 6, the favorable outcomes are the permutations of (3, 3, 0) and (2, 2, 2), along with the permutations of (3, 2, 1). This yields 3 arrangements plus 1 arrangement plus 6 arrangements, totaling 10 favorable outcomes. The probability is 10/216, which simplifies to 5/108.

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