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Re: Two number cubes with six faces numbered with the integers f
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21 Aug 2018, 18:25
Explanation
First think about the prime numbers less than 12, the maximum sum of the numbers on the cube. These primes are 2, 3, 5, 7, 11.
The probability of rolling 2, 3, 5, 7, or 11 is equal to the number of ways to roll any of these sums divided by the total number of possible rolls. The total number of possible cube rolls is 6 × 6 = 36.
Make a list:
Sum of 2 can happen 1 way: 1 + 1.
Sum of 3 can happen 2 ways: 1 + 2 or 2 + 1.
Sum of 5 can happen 4 ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1.
Sum of 7 can happen 6 ways: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1.
Sum of 11 can happen 2 ways: 5 + 6, 6 + 5.
That’s a total of 1 + 2 + 4 + 6 + 2 = 15 ways to roll a prime sum.
Thus, the probability is \(\frac{15}{36}=\frac{5}{12}\).