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Let omega be a complex cube root unity w...

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`

A

`1/18`

B

`1/9`

C

`2/9`

D

`1/36`

Text Solution

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The correct Answer is:
C
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