You visited us 1 times! Enjoying our articles? Unlock Full Access!

Intersection of Sets

Let ω be a ...

Question

Let $ω$ be a complex cube root of unity with $ω$ $\neq$ 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 $ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0$ is

\frac{1}{18}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{36}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{2}{9}$
$r_{1}, r_{2}, r_{3} \in {1, 2, 3, 4, 5, 6}$
$r_{1}, r_{2}, r_{3}$ are if the form $3 k, 3 k + 1, 3 k + 2$
Required probability $\frac{3! \times^{2} C_{1} \times^{2} C_{1} \times^{2} C_{1}}{6 \times 6 \times 6} = \frac{6 \times 8}{216} = \frac{2}{9}$

Suggest Corrections

Similar questions

ω

ω \neq 1

r_{1}, r_{2} a n d r_{3}

ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0,

ω

ω \neq 1

r_{1}, r_{2}

r_{3}

ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0

(n_{1}, n_{2}, n_{3})

(1 \leq n_{1}, n_{2}, n_{3} \leq 6)

i^{n_{1}} + i^{n_{2}} + i^{n_{3}} = 1,

i = \sqrt{- 1}

ω

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + ω & 1 + ω^{2} 1 + ω & 1 + ω^{2} & 1 1 + ω^{2} & 1 & 1 + ω \end{matrix} ∣ ∣ ∣ ∣ ∣ =

ω

(1 - ω + ω^{2})^{6} + (1 - ω^{2} + ω)^{6} =

Join BYJU'S Learning Program