I'm not sure what sort of diagram your teacher wants.
But here's the way to get the answer.
There are 6×6×6 = 216 ways the dice can land. So the
denominator of the desired probability is 216.
Case 1: There are 3 basic ways the three throws can all be
different and the sum be 9. They are
1. 1,2,6
2. 1,3,5
3. 2,3,4
Each of those can be arranged 3!=6 ways,
So there are 3×6=18 ways the three throws can all be
different and the sum be 9.
Case 2. There are two basic ways exactly two of the dice can
be the same and the sum be 9. They are
1. 1,4,4
2, 2,2,5
Each of those can be arranged 3!/2! = 6/2 = 3 ways,
So there are 2×3=6 ways exactly 2 of the dice can be the
same and the sum be 9.
Case 3. There is only 1 way that all three dice can be the same
and the sum be 9. That 1 way is
1. 3,3,3
So there are 18+6+1 = 25 ways the sum can be 9.
So the probability of getting a sum of 9 is 25 ways out of 216,
or 25/216 or about 11.6 of the time.
Edwin
