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# Let $w$ be a complex cube root of unity with $W \neq 1$. 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 $W^{r_1}+W^{r_2}+W^{r_3}=0$ is

$\begin {array} {1 1} (A)\;\large\frac{1}{18} & \quad (B)\;\large\frac{1}{9} \\ (C)\;\large\frac{2}{9} & \quad (D)\;\large\frac{1}{36} \end {array}$

Can you answer this question?

A dice is thrown thrice $n(s) = 6 \times 6 \times \times 6$
Favourable events
$W^{r_1}+ W^{r_2}+ W^{r_3}=0$
$( r_1, r_2, r_3)$ are ordered 3 triples which can take values.
$(1,2,3)\: (1,5,3)\: (4,2,3)\: (4,5,3)$
$(1,2,6)\: (1,5,6)\: (4,2,6)\: (4,5,6)$
i.e, 8 ordered pairs and each can be arranged in 3! ways.
$3! = 6$
$n(E)=8 \times 6$
$P(E)=\large\frac{8 \times 6}{6 \times 6\times 6}$
$= \large\frac{2}{9}$
answered Jan 15, 2014