Let \(\alpha\) be a root of \(x^2+x+\mathrm{I}=0\) and suppose that a fair die is thrown 3 times. If \(a,…
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
\(\frac{2}{36}\)
\(\frac{1}{27}\)
\(\frac{1}{72}\)
\(\frac{2}{9}\)
Solution
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 \(=\frac{6 \times 4 \times 2}{6 \times 6 \times 6}=\frac{8}{36}=\frac{2}{9}\)