Correct Answer - C
Clearly,
Total number of elementary events =6×6×6=216
Clearly, wr1+wr2+wr3=0
, if one of r1,r2 and r3
takes values from the set {3,6}, other takes values from the set {1,4} and the third takes values from the set {2,5}. The total number of these ways is (.2C1×.2C1×.2C1)×3!
So, favourable number of elementary events
={.2C1×.2C1×.2C1)×3≠48
Hence, required probability =
48
216
=
2
9