Let w be a complex cube root of unity with \omega \neq 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that \left(\omega\right)^{r_{1}} + \left(\omega\right)^{r_{2}} + \left(\omega\right)^{r

Engineering

Mathematics

Conditional Probability

Question

Let w be a complex cube root of unity with $ω \neq$ 1. A fair die is thrown three times. If r₁, r₂ and r₃ are the numbers obtained on the die, then the probability that $ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0$ is

$\frac{1}{18}$

$\frac{1}{9}$

$\frac{2}{9}$

$\frac{1}{36}$

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n(S) = 6 · 6 · 6 = 216

n(A) = 6 · 4 · 2 = 48 1 2 3 4 5 6

$∴ P (E) = \frac{48}{216}$ 1, 4 $\to_{}^{}$ w ; 2, 5 $\to_{}^{}$ w² ; 3, 6 $\to_{}^{}$ 1

$= \frac{2}{9}$

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