Let ω be a complex cube root of unity with ω≠1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1+ωr2+ωr3=0 is

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1/18

1/9

2/9

1/36

answer is C.

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Detailed Solution

r1,r2,r3∈{1,2,3,4,5,6}r1,r2,r3 are of the form 3k,3k+1,3k+2 Required probability =3!×2C1×2C1×2C16×6×6=6×8216=29

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