Hi Reecy: Ok-- you have to use some logic here. Begin with what you know. The problem tells us "An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides"
AND Event A : The sum is greater than 8
So start with what your possible die(dice) rolls are--make a table
Die roll #1 Die roll #2 Total (roll#1 + roll #2)
1 (1,2,3,4,5,6) (all of these will be less than 8--so they don't count)
2 (1,2,3,4,5,6) Only 2+6 gets you to 8, and the sum has to be GREATER than 8
3 6 9
4 5 9 (everything below 5 doesn't add up to more than 8)
4 6 10
5 4 9
5 5 10
5 6 11
6 3 9
6 4 10
6 5 11
6 6 12
So now you see that the choices you have are the numbers 9, 10, 11 and 12. Which of these numbers satisfies the second condition??
.Event B : The sum is divisible by 4
Only one of the above: Only 12 can be evenly divided by 4. Therefore, the dice that were thrown had to be a 6 and a 6!
