JEE Advanced · Mathematics · 32. Probability
Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\). 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 \(\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0\) is
- A
\(\frac{1}{18}\)
- B
\(\frac{1}{9}\)
- C
\(\frac{2}{9}\)
- D
\(\frac{1}{36}\)
Answer & Solution
Correct Answer
(C)
\(\frac{2}{9}\)
Step-by-step Solution
Detailed explanation
Sample space A dice is thrown thrice, \(n(s)=6 \times 6 \times 6\).
Favorable events \(\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0\) ie, \(\left(r_1, r_2, r_3\right)\) are ordered 3-triples which can take values, \(\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}\) ie, 8 ordered pairs and each can be arranged in 3 ! ways \(=6\)
\[
\begin{array}{ll}
\therefore & n(E)=8 \times 6 \\
\Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}
\end{array}
\]
Favorable events \(\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0\) ie, \(\left(r_1, r_2, r_3\right)\) are ordered 3-triples which can take values, \(\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}\) ie, 8 ordered pairs and each can be arranged in 3 ! ways \(=6\)
\[
\begin{array}{ll}
\therefore & n(E)=8 \times 6 \\
\Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}
\end{array}
\]
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