Question 1062142
Three fair six-sided number cubes, each with faces labeled 0,2,4,6,8,10 are all rolled. What is the probability that the sum of the numbers rolled is greater than 20? Express your answer with a common fraction.
<pre>This is much easier than you might think.
Total number of events: {{{6^3 = 216}}}
The numbers you're looking for are:
<b><u>Doubles:</b></u>
 6, 8,  8
 6, 6, 10
 8, 8, 10
2, 10, 10
4, 10, 10
6, 10, 10
8, 10, 10</pre><pre>Number of ways ONE of these doubles appears in the 216 events: {{{" "[3]P[3]/2! = 6/2 = 3}}}
Now, since there are 7 DOUBLES, the number of ways those 7 doubles appear in the 216 events is: 7(3), or 21 times

<b><u>Triplicates:</b></u>
 8,  8,  8
10, 10, 10
These triple numbers ONLY appear ONCE in the 216 events, and since there are 2 of them, then they appear in 2 of the 216 events 

<b><u>Distincts:</b></u>
4, 8, 10
6, 8, 10
To find the number of times each of the above numbers appears in the 216 events, we calculate: {{{" "[3]P[3] = 6}}}, and since there are 2 such sets, we get: 2(6), or 12
 
We now see that the number of numbers that appears with a sum > 20 is: 21 + 2 + 12.
So, the probability that the sum of the numbers rolled is greater than 20 = {{{highlight_green(matrix(1,3, (21 + 2 + 12)/216, or, 35/216))}}}