Question 1157684
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<pre>

The entire space of events has  6*6*6 = 216 elements/(events) with the probability {{{1/216}}}  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 = {{{206/216}}} = {{{103/108}}} = 0.9537 = 95.37%.    <U>ANSWER</U>
</pre>

Solved.