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Let omega be a complex cube root of unit...

Let `omega` be a complex cube root of unity with `omega ne 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 `omega^(r_(1)) + omega^(r_(2)) + omega^(r_(3)) = 0` is

A

`1//18`

B

`1//9`

C

`2//9`

D

`1//36`

Text Solution

Verified by Experts

The correct Answer is:
C
Sample space A dice is thrown thrice ,`n(s)=6xx6xx6`
Fovoratble wvents ` omega^(r1)+omega^(r2)+omega^(r3)=0`
i.e `(r_1,r_2,r_3)` are ordered 3 triples which can take values `{:((1,2,3)", "(1,5,3)", "(4,2,3)", "(4,5,3)),((1,2,6)", "(1,5,6)", "(4,2,6)", "(4,5,6)):}}` i.e 8 ordered pairs and each can be arranged in 3! ways =6
`therefore n(E)=8xx6 rArr P(E)=(8xx6)/(6xx6xx6)=2/9`
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