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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

Verified by Experts

The correct Answer is:
C
`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)`
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Knowledge Check

  • Let omega be a complex cube root of unity with omega!=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

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

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    B
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