Multiple choice

Two die are thrown. Find the probability of the event that the sum of the numbers on their upper faces is multiple of $7$:

  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{1}{9}$
  4. $\displaystyle \frac{2}{7}$
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A Correct answer
Explanation

The possible sums that are multiples of 7 are 7 and 14. Pairs summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totaling 6 outcomes. There are 36 total outcomes, so 6/36 = 1/6.

AI explanation

When two dice are thrown, the total number of possible outcomes is 36. The sums that are a multiple of 7 are 7 and 14, but since the maximum possible sum on two dice is 12, we only consider the sum of 7. The favorable outcomes for a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), totaling 6 pairs. Therefore, the required probability is 6 divided by 36, which simplifies to 1/6, making the result 1/6.

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