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Two fair dice, each with faces numbered 1 through 6, are rolled at the same time. Each die has five exposed faces, which are summed. Express as a common fraction the probability that the least common multiple of the sums of the exposed faces is a multiple of 6.

dgfgrafgdfge111  Nov 22, 2018
 #1
avatar+3478 
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The sum of all six sides is 6*7/2=21.

Now, list all the cases out one by one, in order. Your answer should be 5/12.

tertre  Nov 22, 2018
 #2
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What are the cases? Could you please explain in more detail?

dgfgrafgdfge111  Nov 22, 2018
 #3
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To roll either a total of 6 or 12 you have to roll the following:

 

6+6 = 12

1 + 5 = 6

2 + 4 = 6

3 + 3 = 6

5 + 1 = 6

4 + 2 = 6

I'm assuming that the two dice are distinquishible(such as one is RED and the 2nd GREEN), then you have a total of 6 / 6^2 =6/36 =1/6 probability.

Guest Nov 22, 2018
 #4
avatar+134 
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Actually, tertre is correct. The answer is 5/12. But what are the cases?

dgfgrafgdfge111  Nov 22, 2018
 #6
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\(\left( \begin{array}{cc} \{1,2,3,4,5\} & \{1,2,3,4,6\} \\ \{1,2,3,4,5\} & \{1,2,4,5,6\} \\ \{1,2,3,4,5\} & \{2,3,4,5,6\} \\ \{1,2,3,4,6\} & \{1,2,3,4,5\} \\ \{1,2,3,4,6\} & \{1,2,4,5,6\} \\ \{1,2,3,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,2,3,4,5\} \\ \{1,2,4,5,6\} & \{1,2,3,4,6\} \\ \{1,2,4,5,6\} & \{1,2,3,5,6\} \\ \{1,2,4,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,3,4,5,6\} \\ \{1,2,4,5,6\} & \{2,3,4,5,6\} \\ \{1,3,4,5,6\} & \{1,2,4,5,6\} \\ \{2,3,4,5,6\} & \{1,2,3,4,5\} \\ \{2,3,4,5,6\} & \{1,2,4,5,6\} \\ \end{array} \right)\)

Rom  Nov 23, 2018
 #5
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+1

See the detailed answer here. Hopefully, you can understand it:

 

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.388792.html

Guest Nov 23, 2018
 #7
avatar+789 
0

here

PartialMathematician  Nov 24, 2018

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