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,

Question

Let ω be a complex cube root unity with ω≠1. A fair die is thrown three times. If r1,r2andr3 are the numbers obtained on the die, then the probability that ωr1+ωr2+ωr3=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 Select the correct answer from above options

Answer 1

Correct Answer - C Clearly, Total number of elementary events =6×6×6=216 Clearly, wr1+wr2+wr3=0 , if one of r1,r2 and r3 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 (.2C1×.2C1×.2C1)×3! So, favourable number of elementary events ={.2C1×.2C1×.2C1)×3≠48 Hence, required probability = 48 216 = 2 9