Question 1157684
Three dice are tossed. Find the probability of rolling a sum greater than 5.
<pre>
We find the probability of the complement and subtract from 1.

{{{matrix(1,7,
P(matrix(4,1,sum,greater,than,5)),"",""="","",1,""-"",P(matrix(5,1,sum,NOT,greater,than,5)))}}} 

There are only 10 possible rolls with sums that are NOT greater than 5.
They are these:

 1.  (1,1,1)
 2.  (1,1,2)
 3.  (1,1,3)
 4.  (1,2,1)
 5.  (1,2,2)
 6.  (1,3,1)
 7.  (2,1,1)
 8.  (2,1,2)
 9.  (2,2,1)
10.  (3,1,1)

So the numerator (before reducing) of the probability of the complement event
is 10.

We calculate the number of ways the dice can land.

The first one can land any of 6 ways.
For every way the first one can land, the second one can land 6 ways.
So the first two can land 6∙6 or 36 ways.
For every way the first two can land, the third one can land 6 ways.
So there are 6∙6∙6 = 36∙6 = 216 ways the three dice can land.

So the denominator (before reducing) of the probability of the complement event
is 216.

{{{matrix(1,15,
P(matrix(4,1,sum,greater,than,5)),"",""="","",1,""-"",P(matrix(5,1,sum,NOT,greater,than,5)), "", ""="", "", 10/216,"",""="","",5/108 )}}}

Edwin</pre>