Let `omega` be a complex cube root unity with `omega!=1.` A fair die is thrown three times. If `r_1, r_2a n dr_3` are the numbers obtained on the die, then the probability that `omega^(r1)+omega^(r2)+omega^(r3)=0` is

Options

A. `(1)/(18)`
B. `(1)/(9)`
C. `(2)/(9)`
D. `(1)/(36)`

Answer

C

Explanation

A die is thrown thrice, `n(S) =6xx6xx6`
Favourable outcomes of `omega^(r_(1))+omega^(r_(2))+omega^(r_(3))=0`
Clearly `(r_(1),r_(2),r_(3))` are ordered triplets which can take values,
`{:("(1,2,3), (1,5,3), (4,2,3), (4,5,3)"),("(1,2,6), (1,5,6), (4,2,6), (4,5,6)"):}}`
8 ordered triplets and each can be arranged in 3! ways = 6
Let E is an event that `omega^(r_(1))+omega^(r_(2))+omega^(r_(3))=0`
`n(E )= 8xx6 rArr P(E )=(8xx6)/(6xx6xx6)=(2)/(9)`