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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2010 · Paper 1 · Mathematics · Probability

Let ω be a complex cube root of unity with ω ≠ 1 . A fair die is...

Let ω\omega be a complex cube root of unity with ω≠1\omega \neq 1. A fair die is thrown three times. If r1,r2r_1, r_2 and r3r_3 are the numbers obtained on the die, then the probability that ωr1+ωr2+ωr3=0\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0 is

Options

  1. A118\frac{1}{18}
  2. B19\frac{1}{9}
  3. C29\frac{2}{9}
  4. D136\frac{1}{36}

Correct answer

C. 29\frac{2}{9}

Step-by-step solution

Sample space A dice is thrown thrice, n(s)=6×6×6n(s)=6 \times 6 \times 6. Favorable events ωr1+ωr2+ωr3=0\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0 ie, (r1,r2,r3)\left(r_1, r_2, r_3\right) are ordered 3-triples which can take values, \left.(1,2,3),(1,5,3),(4,2,3),(4,5,3)(1,2,6),(1,5,6),(4,2,6),(4,5,6)\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array} ie, 8 ordered pairs and each can be arranged in 3 ! ways =6=6∴n(E)=8×6⇒P(E)=8×66×6×6=29\begin{array}{ll} \therefore& n(E)=8 \times6 \\ \Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9} \end{array}$
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