2)
a)
Number of ways three dice show different outcome = 6*5*4 =120
x1 + x2 +x3 = 10 where xi are all different and 1 <= xi <=6
(1,3,6),(1,4,5),(1,5,4),(1,6,3),
,(2,3,5),(2,5,3),
(3,1,6),(3,2,5),(3,5,2),(3,6,1)
(4,1,5),(4,5,1)
(5,1,4),(5,2,3),(5,3,2),(5,4,1)
(6,1,3),(6,3,1)
total number of ways = 18
hence required probability = 18/120 = 3/20
option A) is correct
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