Question

Let w be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If ${{\text{r}}_1},{\text{ }}{{\text{r}}_2}{\text{ and }}{{\text{r}}_3}$ 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
${\text{A}}{\text{. }}\dfrac{1}{{18}} \\\ {\text{B}}{\text{. }}\dfrac{1}{9} \\\ {\text{C}}{\text{. }}\dfrac{2}{9} \\\ {\text{D}}{\text{. }}\dfrac{1}{{36}} \\\$

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 ${\omega ^{{{\text{r}}_1}}} + {\omega ^{{{\text{r}}_2}}} + {\omega ^{{{\text{r}}_3}}} = 0$ ,
$\Rightarrow {\omega ^{{\text{3k}}}} + {\omega ^{{\text{3k + 1}}}} + {\omega ^{{\text{3k + 2}}}} \\\ \Rightarrow {\omega ^{{\text{3k}}}}\left( {1 + \omega + {\omega ^2}} \right) \\\ \Rightarrow 0 \\\$ --- 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 $\dfrac{{{\text{number of favorable ways}}}}{{{\text{total number of possibilities}}}}$
⟹Probability = $\dfrac{{48}}{{216}} = \dfrac{2}{9}$
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.