Arun
Last Activity: 7 Years ago
Dear Sohan
Number of triplets favourable to the event that the sum of the faces of three dice is 15 is (6,6,3),(6,5,4),(6,4,5),(6,3,6),(5,6,4),(5,5,5),(5,4,6),(4,6,5),(4,5,6) and (3,6,6). Hence, the required probability = 10 * 1/216 = 5/108
Regards
Arun (askIITians forum expert)