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
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`