omega is a complex cube root of unity. When a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The total number of outcomes when a die is thrown $3$ times is $6 	imes 6 	imes 6 = 216$. Given the condition $\omega^{\beta_1} + \omega^{\beta_

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(D) The total number of outcomes when a die is thrown $3$ times is $6 	imes 6 	imes 6 = 216$. Given the condition $\omega^{\beta_1} + \omega^{\beta_2} = -\omega^{\beta_3}$, we can rewrite this as $\omega^{\beta_1} + \omega^{\beta_2} + \omega^{\beta_3} = 0$. Since $\omega$ is a complex cube root of unity, $\omega^n$ can take values $\omega, \omega^2, 1$ depending on whether $n \equiv 1, 2, 0 \pmod{3}$. For the sum to be zero, the values ${\omega^{\beta_1}, \omega^{\beta_2}, \omega^{\beta_3}}$ must be a permutation of ${1, \omega, \omega^2}$. This means one of $\beta_1, \beta_2, \beta_3$ must be of the form $3k$, one of the form $3k+1$, and one of the form $3k+2$. In the set ${1, 2, 3, 4, 5, 6}$, there are two numbers of each type: Type $0$ $(n \equiv 0 \pmod{3})$: ${3, 6}$ Type $1$ $(n \equiv 1 \pmod{3})$: ${1, 4}$ Type $2$ $(n \equiv 2 \pmod{3})$: ${2, 5}$ The number of ways to choose one number from each set is $2 	imes 2 	imes 2 = 8$. Since the order of $\beta_1, \beta_2, \beta_3$ matters, we multiply by $3! = 6$. Favourable outcomes $= 8 	imes 6 = 48$. Probability $= \frac{48}{216} = \frac{2}{9}$.

$\omega$ is a complex cube root of unity. When an unbiased die is thrown $3$ times, if $\beta_1, \beta_2, \beta_3$ are the numbers appeared on the die, then the probability that $\beta_1, \beta_2$ and $\beta_3$ satisfy $\omega^{\beta_1}+\omega^{\beta_2}=-\omega^{\beta_3}$ is

Similar Questions

Three squares of a chessboard are selected at random. The probability of selecting two squares of one colour and the other of a different colour is equal to

The numbers $2, 3, 5, 7, 11, 13$ are written on six distinct paper chits. If $3$ of them are chosen at random, then the probability that the sum of the numbers on the obtained chits is divisible by $3$ is:

An urn contains $8$ red and $5$ white balls. Three balls are drawn at random. Then, the probability that balls of both colours are drawn is

There are $8$ boys and $7$ girls in a classroom. If the names of all those children are written on paper slips and $3$ slips are drawn at random from them,then the probability of getting the names of one boy and two girls or one girl and two boys is

$A$ bag contains $6$ red,$4$ white,and $8$ blue balls. If three balls are drawn at random,then the probability that $2$ are white and $1$ is red is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module