Probability Ques 1
- 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
(2010)
(a) $1 / 18$
(b) $1 / 9$
(c) $2 / 9$
(d) $1 / 36$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Sample space A dice is thrown thrice, $n(s)=6 \times 6 \times 6$.
Favorable events $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$
i.e. $\left(r_1, r_2, r_3\right)$ are ordered 3 triples which can take values,
$\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}$
i.e. $8$ ordered pairs and each can be arranged in $3$ ! ways $=6$
$ \therefore \quad n(E)=8 \times 6 \Rightarrow P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9} $
BETA
