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Two number cubes with six faces numbered with the integers f
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05 Aug 2018, 14:34
05 Aug 2018, 14:34
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Question Stats:
43% (01:46) correct
56% (04:41) wrong based on 46 sessions
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Two number cubes with six faces numbered with the integers from 1 through 6 are tossed. What is the probability that the sum of the exposed faces on the cubes is a prime number?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{5}{12}\)
Re: Two number cubes with six faces numbered with the integers f
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11 Aug 2018, 12:33
11 Aug 2018, 12:33
2
We note which numbers between 1 and 12 are prime numbers (12 is the highest sum two dice can combine for). 2,3,5,7,11.
One combination yields sum 2
Two combinations yield sum 3
Four combinations yield sum 5
Six combinations yield sum 7
Two combinations yield sum 11
So in total 15 combinations yield a prime sum. There are 36 possible ways the dice can land if you roll two dice (6 faces, 2 dice, 6^2=36), and so the probability is 15/36, or 5/12 simplified.
One combination yields sum 2
Two combinations yield sum 3
Four combinations yield sum 5
Six combinations yield sum 7
Two combinations yield sum 11
So in total 15 combinations yield a prime sum. There are 36 possible ways the dice can land if you roll two dice (6 faces, 2 dice, 6^2=36), and so the probability is 15/36, or 5/12 simplified.
Re: Two number cubes with six faces numbered with the integers f
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21 Aug 2018, 18:25
21 Aug 2018, 18:25
4
Expert Reply
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 \(\frac{15}{36}=\frac{5}{12}\).
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 \(\frac{15}{36}=\frac{5}{12}\).
