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.