Question

Let `omega` be a complex cube root unity with `omega!=1.` A fair die is thrown three times. If `r_1, r_2a n dr_3` are the numbers obtained on the die, then the probability that `omega^(r1)+omega^(r2)+omega^(r3)=0` is `1//18` b. `1//9` c. `2//9` d. `1//36`
A. `(1)/(18)`
B. `(1)/(9)`
C. `(2)/(9)`
D. `(1)/(36)`

Answer 1

Correct Answer - C
Clearly,
Total number of elementary events `=6xx6xx6=216`
Clearly, `w^(r_(1))+w^(r_(2))+w^(r_(3))=0`, if one of `r_(1), r_(2) " and " r_(3)` takes values from the set {3,6}, other takes values from the set {1,4} and the third takes values from the set {2,5}. The total number of these ways is `(.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!`
So, favourable number of elementary events
`={.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!=48`
Hence, required probability `=(48)/(216)=(2)/(9)`