MCQ
A six faced fair dice is thrown until $2$ comes, then the probability that $2$ comes in even number of trials is (dice having six faces numbered $1, 2, 3, 4, 5$ and $6$)
- A$\frac {1}{6}$
- B$\frac {5}{6}$
- C$\frac {6}{11}$
- ✓$\frac {5}{11}$
$\therefore \quad \mathrm{P}(2 \text { comes in even trial })$
${=\mathrm{P}(\bar{z} \mathrm{z} \text { or } \bar{z} \bar{z} \bar{z} z \text { or } \ldots \ldots \ldots \infty)} $
${=\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\ldots \ldots \infty} $
${=\frac{5}{6} \times \frac{1}{6}} $
${=\frac{\frac{5}{6} \times \frac{1}{6}}{1-\left(\frac{5}{6}\right)^{2}}=\frac{5}{11}}$
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$a+b+c $$ =x $ ; $a+b \omega+c \omega^2 $$ =y $ ; $a+b \omega^2+c \omega $$ =z .$
Then the value of $\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2}$ is