Question

Let $\alpha$ be a root of $x^2+x+1=0$ and suppose that a fair die is thrown $3$ times. If $a,b$ and $c$ are the numbers shown on the die, then the probability that ${\alpha }^a+{\alpha }^b+{\alpha }^c=0$, is

• A. $\frac{2}{36}$
• B. $\frac{1}{27}$
• C. $\frac{1}{72}$
• D. $\frac{2}{9}$

Answer 1

Answer: (D)

Total numbers of ways for $(a,b,c)=6\times 6\times 6$
Here, $\omega ,{\omega }^2$
${\omega }^3=1$
${\omega }^4=\omega$
${\omega }^5={\omega }^2$
and ${\omega }^6=1$
Since, ${\omega }^2,\omega$ are roots of $x^2+x+1=0$
$\because {\omega }^a+{\omega }^b+{\omega }^c=0$
Now, suitable values for $(a,b,c)=6\times 4\times 2$
$\therefore$ Required probability $=\frac{6\times 4\times 2}{6\times 6\times 6}=\frac{8}{36}=\frac{2}{9}$