IIT-JEE 2010 · Paper 1 · Mathematics Q31

← IIT-JEE 2010 · Paper 1

MathematicsSingle correct+3 / -1Question 31 of 84

Let $\omega$ be a complex cube root of unity with $\omega\ne1$ . 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

\dfrac1{18}

\dfrac19

\dfrac29

\dfrac1{36}

Answer and worked solution

Official answer: C

When is the sum zero? $\omega^r$ depends only on $r\bmod3$ , and $1+\omega+\omega^2=0$ is the only way three cube roots of unity add to zero. So $r_1,r_2,r_3$ must leave remainders 0, 1, 2 in some order.

Count. Each remainder class has two faces ({3,6}, {1,4}, {2,5}): $3!\times2^3=48$ outcomes out of 216.

Probability. $\frac{48}{216}=\frac29$ .

(C)

Topics:Classical probability Roots of unity

← Previous question Next question →

IIT-JEE 2010 · Paper 1, Mathematics question 31