TS ICET 2017 Question Paper with Solutions for online practice

Total number of outcomes $=6^3$
In this case we just need to count possibilities for two dice because the value of the third dice automatically gets fixed. For two dice, the sum can be anywhere between 8 and 12 (as the maximum on the third dice can only be 6 and also 13 is not possible on two dice).

Third dice	First Second dice	Number of ways
6	(6,2),(5,3),(4,4)	5
5	(6,3),(5,4)	4
4	(6,4),(5,5)	3
3	(6,5)	2
2	(6,6)	1
		Total = 15

∴ Required probability $=\frac{15}{6\times 6\times 6}=\frac{5}{72}$