Let ω be a complex cube root of unity with ω ≠ 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 r1 +ω r2 +ω r3=0 , is

Question

Let ω be a complex cube root of unity with ω ≠ 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 r1 +ω r2 +ω r3=0 , is (A) 1/18 (B) 1/9 (C) 2/9 (D) 1/36

Tardigrade Admin · Accepted Answer

Sample space A dice is thrown thrice, n(s)=6 × 6 × 6. Favorable events ωr1+ωr2+ωr3=0 i.e. (r1, r2, r3) are ordered 3 triples which can take values, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/9de386daa7aa30bfe9aa5489826f40df-.png" /> and each can be arranged in 3 ! ways =6 ∴ n(E)=8 × 6 ⇒ P(E)=(8 × 6/6 × 6 × 6)=(2/9)

Q. Let ω be a complex cube root of unity with ω=1. A fair die is thrown three times. If r1​,r2​ and r3​ are the numbers obtained on the die, then the probability that ωr1​+ωr2​+ωr3​=0, is

1/18

1/9

2/9

1/36

Sample space A dice is thrown thrice, n(s)=6×6×6. Favorable events ωr1​+ωr2​+ωr3​=0 i.e. (r1​,r2​,r3​) are ordered 3 triples which can take values, and each can be arranged in 3! ways =6 ∴n(E)=8×6⇒P(E)=6×6×68×6​=92​

Q. 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