Let $\omega$ be a complex cube root of unity with $\omega\
eq 1$. 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

Question

Let $\omega$ be a complex cube root of unity with $\omega\ 
eq 1$. 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

Dr. Abubacker · Accepted Answer

(c)
A dice is thrown thrice, $n(S) = 6 	imes 6 	imes 6$
Favorable events of $\omega^{r_{1}} + \omega^{r_{2}} + \omega^{r_{3}} = 0$
$ie,\ \left( r_{1},\ r_{2},\ r_{3} ight)$ are ordered triplets which can take values,
$\left. \ \begin{array}{r}
(1,\ 2,\ 3),\ (1,\ 5,\ 3),\ (4,\ 2,\ 3),\ (4,\ 5,\ 3) \
(1,\ 2,\ 6),\ (1,\ 5,\ 6),\ (4,\ 2,\ 6),\ (4,\ 5,\ 6)
\end{array} ight\}\$
$ie,\ 8$ ordered triplets and each can be arranged in 3! ways $= 6$
$	herefore n(E) = 8 	imes 6$
$\Rightarrow P(E) = \frac{8 	imes 6}{6 	imes 6 	imes 6}$
$= \frac{2}{9}$

Q. Let $\omega$ be a complex cube root of unity with $\omega\ \neq 1$. 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

Explanation

Reviewed by Dr. Abubacker

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