Question

Let w be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If ${\text{r}}_1,{\text{r}}_2\text{ and }{\text{r}}_3$ are the numbers obtained on the die then the probability that ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$ is
 $$\begin{gathered} \text{A}\text{. }{\displaystyle \frac{1}{18}} \\ \text{B}\text{. }{\displaystyle \frac{1}{9}} \\ \text{C}\text{. }{\displaystyle \frac{2}{9}} \\ \text{D}\text{. }{\displaystyle \frac{1}{36}} \end{gathered}$$

Answer 1

Hint: In order to find the probability we use the condition of cube roots of unity, then we compute the possibilities of ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$ and then apply the formula of probability to find the answer.
 We will use one property of cube root of unity i.e. $1+\omega +{\omega }^2=0$

Complete step-by-step answer:
Given Data, ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$
The total number of outcomes when a fair die is rolled 3 times is 6 × 6 × 6 = 216.
Cube root of unity is bound by a condition:- $1+\omega +{\omega }^2=0$
The numbers of the die are 1, 2, 3, 4, 5 and 6.
The numbers are of the form 3k, 3k+1 and 3k+2, i.e.
3k --- 3 and 6
3k+1 --- 1 and 4
3k+2 --- 2 and 5
Let ${\text{r}}_1$ ϵ 3k, ${\text{r}}_2$ ϵ 3k+1 and ${\text{r}}_3$ ϵ 3k+2
Now ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$ ,
 $$\begin{gathered} \Rightarrow {\omega }^{\text{3k}}+{\omega }^{\text{3k + 1}}+{\omega }^{\text{3k + 2}} \\ \Rightarrow {\omega }^{\text{3k}}\left(1+\omega +{\omega }^2\right) \\ \Rightarrow 0 \end{gathered}$$ --- As we know $1+\omega +{\omega }^2=0$
Hence ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$ is true, only if ${\text{r}}_1$ ϵ 3k, ${\text{r}}_2$ ϵ 3k+1 and ${\text{r}}_3$ ϵ 3k+2.
Now the number of favorable outcomes =
We can arrange each of the elements from ${\text{r}}_1$ , ${\text{r}}_2$ and ${\text{r}}_3$ in ${}^2{\text{C}}_1$ × ${}^2{\text{C}}_1$ × ${}^2{\text{C}}_1$ ways.
And we can arrange ${\text{r}}_1$ , ${\text{r}}_2$ and ${\text{r}}_3$ in 3! Ways.
(As we know we can arrange n terms in n! ways and n! = n (n-1) (n-2)…… (n - (n-1))).
Hence the number of favorable ways = 3! × ${}^2{\text{C}}_1$ × ${}^2{\text{C}}_1$ × ${}^2{\text{C}}_1$
                                                                   = 6 × 2 × 2 × 2
                                                                   = 48.
Hence the probability that ${\omega }^{{\text{r}}_1}+{\omega }^{{\text{r}}_2}+{\omega }^{{\text{r}}_3}=0$ is ${\displaystyle \frac{\text{number of favorable ways}}{\text{total number of possibilities}}}$
⟹Probability = ${\displaystyle \frac{48}{216}}={\displaystyle \frac{2}{9}}$
Hence Option C is the correct answer.

Note – In order to solve this type of problems the key is to have enough knowledge in the cube roots of unity and its condition. A fair dice has 6 possible outcomes when rolled once. It is important to identify that picking a term from each of ${\text{r}}_1$ , ${\text{r}}_2$ and ${\text{r}}_3$ is a combination but not a permutation, also while finding the favorable possibilities we must remember to consider the possibilities of arranging ${\text{r}}_1$ , ${\text{r}}_2$ and ${\text{r}}_3$ respectively.