Let $\alpha$ be a root of $x^2+x+\mathrm{I}=0$ 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

Question

Let $\alpha$ be a root of $x^2+x+\mathrm{I}=0$ 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, $\frac{2}{36}$, $\frac{1}{27}$, $\frac{1}{72}$, $\frac{2}{9}$

MARKS App · Accepted Answer

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

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