Complex Probability

Probability Level 4

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ .A fair die is thrown three times. If $r_1$ , $r_2$ and $r_3$ are the numbers obtained on the die, then the probability that is $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is

\dfrac{2}{9}

\dfrac{1}{9}

\dfrac{1}{18}

None of these

\dfrac{1}{36}

1 solution

Pulkit Gupta
Dec 7, 2015

Let $r_{1}$ be 3m , $r_{2}$ be 3n+1 & $r_{3}$ be 3p+1 . Here, m can take values 1,2 ; n can take values 0,1 ; p can take values 0,1. Note that m CANNOT be zero.

Also note that the order in which these values appear does not matter.

By the above facts, favorable cases are 2 * 2 * 2 * 3!

The sample space consists of 6 * 6 * 6 or 216 values.

Complex Probability

1 solution

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