Question

Let $ \omega $ be the 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). $ \dfrac{1}{18} $
b). $ \dfrac{1}{9} $
c). $ \dfrac{2}{9} $
d). $ \dfrac{1}{36} $

Answer 1

Hint:
To find the probability of the above question we will use the condition of the cube root of the unity i.e. $ 1+\omega +{{\omega }^{2}}=0 $ and $ {{\omega }^{3}}=1 $ , and then we will apply the formula of the probability to get the required answer.

Complete step by step answer:
We know that from the question that a fair die is rolled 3 times. So, the total number of possible outcomes is $ 6\times 6\times 6=216 $ because each time there are three possible outcomes.
Also, we know that $ \omega $ is the complex cube root of unity. So, we will have $ 1+\omega +{{\omega }^{2}}=0 $ and also $ {{\omega }^{3}}=1 $ .
We know that the total number of possible outcomes on the die is 1, 2, 3, 4, 5, 6.
Now, we know from the question that we have to compute the probability of $ {{\omega }^{{{r}_{1}}}}+{{\omega }^{{{r}_{2}}}}+{{\omega }^{{{r}_{3}}}}=0 $ .
It is zero only when $ {{r}_{1}},{{r}_{2}},{{r}_{3}} $ is of the form 3k, 3k+1 and 3k+2 respectively because $ {{\omega }^{3}}=1 $ so,
 $ {{\omega }^{3k}}={{1}^{k}}=1 $ , $ {{\omega }^{3k+1}}={{\omega }^{3k}}\times \omega =\omega $ and $ {{\omega }^{3k+2}}={{\omega }^{2}} $
So, we will get $ 1+\omega +{{\omega }^{2}}=0 $ .
Hence, from 3k = 3 and 6, we get $ {{\omega }^{0}}=1 $
Similarly, from 3k + 1 = 1 and 4, we get $ {{\omega }^{1}}=\omega $
And, from 3k + 2 = 2 and 5, we get $ {{\omega }^{2}} $
 So, total number of possible ways is which we can arrange each element from $ {{r}_{1}} $ , $ {{r}_{2}} $ and $ {{r}_{3}} $ is equal to $ {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}} $ .
Also, we can arrange $ {{r}_{1}} $ , $ {{r}_{2}} $ and $ {{r}_{3}} $ among themselves in $ 3! $ number of ways.
So, the total number of favourable outcomes is $ 3!\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}} $ .
So, the probability that $ {{\omega }^{{{r}_{1}}}}+{{\omega }^{{{r}_{2}}}}+{{\omega }^{{{r}_{3}}}}=0 $ is given by $ \dfrac{\text{ T}otal\text{ }number\text{ }of\text{ }favourable\text{ }outcomes}{Total\text{ }number\text{ }of\text{ }possible\text{ }outcomes} $
So, Probability is equal to $ \dfrac{3!\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}}{216} $ = $ \dfrac{6\times 2\times 2\times 2}{216} $ = $ \dfrac{48}{216} $ = $ \dfrac{2}{9} $
Hence, the option (c) is the correct answer. This is our required solution.

Note:
Students are required to note that when we are given cube root of unity in the question then there is much chance that $ 1+\omega +{{\omega }^{2}}=0 $ and $ {{\omega }^{3}}=1 $ is applicable in that question and also students are required to note that they should not miss any combination in this question other they will get the wrong answer.