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  • A fair n(n > 1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is <img alt="\frac{n}{9}" src="https://entrancecorner.oncod

A fair n(n > 1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of
tosses required is \frac{n}{9}, then n is equal to ______.

Option: 1

10


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Mean }=1, \frac{n-1}{n}+2 \frac{1}{n}\left(\frac{n-1}{n}\right)+3\left(\frac{1}{n}\right)^2\left(\frac{n-1}{n}\right) \ldots \\ & \frac{\mathrm{n}}{9}=\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)\left(1+2\left(\frac{1}{n}\right)+3\left(\frac{1}{n}\right)^2 \ldots \ldots \cdot\right) \\ & \frac{\mathrm{n}}{9}=\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)\left(1-\frac{1}{n}\right)^{-2}=\left(\frac{n-1}{n}\right) \cdot \frac{n^2}{(n-1)^2} \\ & \frac{\mathrm{n}}{9}=\frac{\mathrm{n}}{\mathrm{n}-1} \Rightarrow \mathrm{n}=10 \end{aligned}

Posted by

vishal kumar

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