We find the probability of the complement and subtract from 1.

 

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.



Edwin

The entire space of events has  6*6*6 = 216 elements/(events) with the probability   each.


Of them undesired are


    1 event  with the sum of 3   [ the event (1,1,1) ]

    3 events with the sum of 4   [ the events (1,1,2), (1,2,1)  and (2,1,1) ]

    6 events with the sum of 5   [ the events (1,1,3), (1,3,1), (3,1,1),

                                              (1,2,2), (2,1,2), (2,2,1) ].


The rest 216 - 10 = 206 events are favorable.


Therefore, the probability under the question is  P =  =  = 0.9537 = 95.37%.    ANSWER