JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Complex Numbers

Let

ω

be a complex cube root of unity with

ω \neq = 1

. 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

ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0

Answer: (C)

Solution

We have

ω^{3} = 1

and

1 + ω + ω^{2} = 0

.

For

ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0

, we need the exponents modulo 3 to sum appropriately.

Since

r_{i} \in {1, 2, 3, 4, 5, 6}

, we have

r_{i} mod 3 \in {0, 1, 2}

with:

For the sum to be zero, we need one of each residue class (0, 1, 2) modulo 3.

Number of favorable outcomes =

3! \times 2^{3} = 6 \times 8 = 48

Total outcomes =

6^{3} = 216

Probability =

\frac{48}{216} = \frac{2}{9}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option