Let omega be a complex cube a root of u...

Let `omega ` be a complex cube a 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)`

Verified by Experts

The correct Answer is:

C

Similar Questions

Explore conceptually related problems

If w is a complex cube-root of unity then,

If w is a complex cube roots of unity, then arg (i w)+arg (i w^(2))=

If omega is a cube root of unity (1-2 omega+omega^(2))^(6)=

Let omega be a complex cube root of unity with omega!=1a n dP=[p_(i j)] be a nxxn matrix withe p_(i j)=omega^(i+j)dot Then p^2!=O ,when=

If omega is a non real cube root of unity then (a+b)(a+b omega)(a+b omega^(2)) is

If omega is an imaginary cube root of unity than (1+omega-omega^(2))^(7)=

If omega is a complex cube root of unity , then the value of omega^(99) + omega^(100) + omega^(101) is

1, omega, omega^(2) are cube roots of unit x, then their product is

If 1, omega , omega^(2) are the cube roots of unity then (1 + omega) (1 + omega^(2))(1 + omega^(4))(1 + omega^(8)) is equal to

If omega is an imaginary cube root of unity, then (1+omega+omega^(2))^(7) equals :