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 `1//18` b. `1//9` c. `2//9` d. `1//36`

Options

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

Answer

C

Explanation

`r_(1),r_(2), r_(3) in {1, 2, 3, 4, 5, 6}`
`r_(1), r_(2), r_(3)` are of the form 3k, 3k + 1, 3k + 2
Required probability = `(3! xx .^(2)C_(1)xx.^(2)C_(1) xx.^(2)C_(1))/(6xx6xx6) = (6xx8)/(216) = (2)/(9)`