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. 44579
• B. 44570
• C. 44601
• D. Jan-36

Answer 1

Answer: (C)

44601

Answer 2

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. $(r_1,r_2,r_3)$ are ordered 3 triples which can take
values
\begin {array} \ (1,2,3) \\\ (1,2,6) \end {array} \begin {array} \ (1,5,3) \\\ (1,5,6) \end {array} \begin {array} \ (4,2,3) \\\ (4,2,6) \end {array} \begin {array} \ (4,5,3) \\\ (4,5,6) \end {array} \bigg \\} i.e., \ \ 8 \ ordered \ pairs and each can be arranged in 3! ways = 6
'
$\therefore \ \ \ \ \ \ \ \ n(E)=8 \times 6 \ \ \Rightarrow \ \ p(E) =\frac{8 \times 6}{6 \times 6 \times 6} =\frac{2}{9}$