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**+1 **

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

#3**+1 **

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**+1 **

Actually, tertre is correct. The answer is 5/12. But what are the cases?

dgfgrafgdfge111
Nov 22, 2018

#6**+1 **

\(\left( \begin{array}{cc} \{1,2,3,4,5\} & \{1,2,3,4,6\} \\ \{1,2,3,4,5\} & \{1,2,4,5,6\} \\ \{1,2,3,4,5\} & \{2,3,4,5,6\} \\ \{1,2,3,4,6\} & \{1,2,3,4,5\} \\ \{1,2,3,4,6\} & \{1,2,4,5,6\} \\ \{1,2,3,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,2,3,4,5\} \\ \{1,2,4,5,6\} & \{1,2,3,4,6\} \\ \{1,2,4,5,6\} & \{1,2,3,5,6\} \\ \{1,2,4,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,3,4,5,6\} \\ \{1,2,4,5,6\} & \{2,3,4,5,6\} \\ \{1,3,4,5,6\} & \{1,2,4,5,6\} \\ \{2,3,4,5,6\} & \{1,2,3,4,5\} \\ \{2,3,4,5,6\} & \{1,2,4,5,6\} \\ \end{array} \right)\)

Rom
Nov 23, 2018