Probability Question 7
Question: Let $ \omega $ be a complex cube root of unity with $ \omega \ne 1. $ A fair die is thrown three times. If $ r,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
Options:
A) 1/18
B) 1/9
C) 2/9
D) 1/36
Show Answer
Answer:
Correct Answer: C
Solution:
$ r_1,r_2,r_3\in {1,2,3,4,5,6} $
$ r_1,r_2,r_3 $ and of the form $ 3k,3k+1,3k+2 $
Required probability $ =\frac{3!{{\times }^{2}}C_1{{\times }^{2}}C_1{{\times }^{2}}C_1}{6\times 6\times 6}=\frac{6\times 2}{216}=\frac{1}{18}. $
BETA
