Home
Class 12
MATHS
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_2 and r_3` are the numbers obtained on the die, then the probability that `omega^(r1) + omega^(r2) + omega^(r3)=0` is (a) `1/18` (b) `1/9` (c) `2/9` (d) `1/36`

Promotional Banner

Topper's Solved these Questions

  • JEE (ADVANCED ) 2020

    JEE ADVANCED PREVIOUS YEAR|Exercise SECTION 2|6 Videos

  • JEE (ADVANCED ) 2020

    JEE ADVANCED PREVIOUS YEAR|Exercise SECTION 3|6 Videos

  • JEE (ADVANCED) 2020

    JEE ADVANCED PREVIOUS YEAR|Exercise SECTION-3|6 Videos

Similar Questions

Explore conceptually related problems

Let omega be a complex cube root unity with omega!=1. A fair die is thrown three xx.If r_(1),r_(2) and r_(3) are the numbers obtained on the die,then the probability that omega^(r1)+omega^(r2)+omega^(r3)=0 is (a) (1)/(18) (b) (1)/(9)(c)(2)/(9) (d) (1)/(36)

Let omega be a cube of unity . If r_1,r_2,and r_3 be the numbers obtained on the die. Then probability of omega^(r_1)+omega^(r_2)+omega^(r_3)=0 is

Let omega be a complex cube root of unity with 0

If omega ne 1 is a cube root of unity, then 1, omega, omega^(2)