Question

Let $\mathrm{\omega }$ be a complex cube root of unity with $\mathrm{\omega }\ne 1$. A fair die is thrown three times. If $\mathrm{r}1,\mathrm{r}2$ and $\mathrm{r}3$ are the numbers obtained on the die, then the probability that ${\mathrm{\omega }}^{\mathrm{r}1}+{\mathrm{\omega }}^{\mathrm{r}2}+{\mathrm{\omega }}^{\mathrm{r}3}=0$ is

• A. $1/18$
• B. $1/9$
• C. $2/9$
• D. $1/36$

Answer 1

The correct option is C

$2/9$

Determine the probability that ${\mathrm{\omega }}^{\mathrm{r}1}+{\mathrm{\omega }}^{\mathrm{r}2}+{\mathrm{\omega }}^{\mathrm{r}3}=0$

Given, $\mathrm{\omega }$ is the complex cube root of unity, therefore we know that sum of the cube root of unity is equal to zero, that is, $1+\mathrm{\omega }+{\mathrm{\omega }}^2=0...\left(1\right)$

A fair die is thrown three times, hence its sample space or total number of outcomes will be equal to $6\times 6\times 6=216$

The sample space of a fair die is thrown one time is $\left\{1,2,3,4,5,6\right\}$

Let us categorize the sample space into three parts that are,

$$\begin{gathered} 3\mathrm{k}\to \left\{3,6\right\} \\ 3\mathrm{k}+1\to \left\{1,4\right\} \\ 3\mathrm{k}+2\to \left\{2,5\right\}\mathrm{k}=0,1,2 \end{gathered}$$

Now let $\mathrm{r}1\in 3\mathrm{k},\mathrm{r}2\in 3\mathrm{k}+1,\mathrm{r}3\in 3\mathrm{k}+2$, hence

$$\begin{gathered} {\mathrm{\omega }}^{\mathrm{r}1}+{\mathrm{\omega }}^{\mathrm{r}2}+{\mathrm{\omega }}^{\mathrm{r}3}={\mathrm{\omega }}^{3\mathrm{k}}+{\mathrm{\omega }}^{3\mathrm{k}+1}+{\mathrm{\omega }}^{3\mathrm{k}+2} \\ ={\mathrm{\omega }}^{3\mathrm{k}}(1+\mathrm{\omega }+{\mathrm{\omega }}^2) \\ ={\mathrm{\omega }}^{3\mathrm{k}}\times 0[\mathrm{from}(1\left)\right] \\ =0 \end{gathered}$$

So the number of ways in which the $\mathrm{r}1,\mathrm{r}2,\mathrm{r}3$ can arrange is, ${}^{2}C_1\times {}^{2}C_1\times {}^{2}C_1\times 3!=48$

Therefore the probability, $\mathrm{P}=\frac{48}{216}$

$\mathrm{P}=\frac{2}{9}$

Hence, option C is the correct answer,