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_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)`
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY I

    CENGAGE|Exercise JEE Main Previous Year|4 Videos

  • PROBABILITY AND STATISTICS

    CENGAGE|Exercise Question Bank|24 Videos

  • PROBABILITY II

    CENGAGE|Exercise JEE Advanced Previous Year|25 Videos

Similar Questions

Explore conceptually related problems

If omega is the cube root of unity, then then value of (1 - omega) (1 - omega^(2))(1 - omega^(4))(1 - omega^(8)) is

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 n= a. 57 b. 55 c. 58 d. 56

If omega is a cube root of unity and (omega-1)^7=A+Bomega then find the values of A and B

If 1, omega omega^(2) are the cube roots of unity show that (1 + omega^(2))^(3) - (1 + omega)^(3) = 0

If omega is a cube root of unity, then find the value of the following: (1+omega-omega^2)(1-omega+omega^2)

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

If two fair dices are thrown and digits on dices are a and b, then find the probability for which omega^(ab) = 1 , (where omega is a cube root of unity).

If omega is a complex nth root of unity, then underset(r=1)overset(n)(ar+b)omega^(r-1) is equal to

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