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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

  • If omega is a cube root of unity, then the value of (1 - omega + omega^2)^4 + (1 + omega - omega^2)^(4) is ……….

    A
    `-16`
    B
    `0`
    C
    `-32`
    D
    `32`

  • If omega ne 1 is a cubic root of unity and (1+ omega)^(7) = A + B omega , then (A,B) equals

    A
    `(1,0)`
    B
    `(-1,1)`
    C
    `(0,1)`
    D
    `(1,1)`

  • If the cube roots of unity are 1, omega, omega^(2) then the roots of the equation (x-1)^(3)+8=0 , are

    A
    `-1,-1,-1`
    B
    `-1,-1+2omega, -1-2 omega^(2)`
    C
    `-1, 1+2omega, 1+2 omega^(2)`
    D
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