Let `omega` be a complex cube root of unity with `omega ne 1`. A fair die is thrown three times. If `r_(1), r_(2)` and `r_(3)` are the numbers obtained on the die, then the probability that `omega^(r_(1)) + omega^(r_(2)) + omega^(r_(3)) = 0` is

Options

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

Answer

C

Explanation

Clearly,
Total number of elementary events `=6xx6xx6=216`
Clearly, `w^(r_(1))+w^(r_(2))+w^(r_(3))=0`, if one of `r_(1), r_(2) " and " r_(3)` 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 `(.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!`
So, favourable number of elementary events
`={.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!=48`
Hence, required probability `=(48)/(216)=(2)/(9)`