Solution:Number of ways in which one of the face having the number 6 and no two dice show the same number.
(1,2,6), (1,3,6), (1,4,6), (1,5,6), (2,3,6), (2,4,6), (2,5,6), (3,4,6), (3,5,6)..........
Total favourable case = 20+20+20 = 60.
Number of total output when three top faces of three dice shows different number = 6 Γ 5 Γ 4 = 120.
β΄ Required probability = 60/120 = 1/2.