Question

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

• A. 1/18
• B. 1/9
• C. 2/9
• D. 1/36

Answer 1

Answer: (C)

Sample space A dice is thrown thrice, $n(s)=6 \times 6 \times 6$.
Favorable events $\omega^{r_{1}}+\omega^{r_{2}}+\omega^{r_{3}}=0$
i.e. $\left(r_{1}, r_{2}, r_{3}\right)$ are ordered $3$ triples which can take values,

and each can be arranged in $3 !$ ways $=6$
$\therefore n(E)=8 \times 6 \Rightarrow P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}$