A fair faces die is rolled repeatedly until a number less than appears. If the mean of the number of tosses required is then is equal to _______ . [2023]

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Solution

Q.
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 _______ . [2023]

Ans.
(10)

$Mean = 1 \cdot \frac{n - 1}{n} + 2 \frac{1}{n} (\frac{n - 1}{n}) + 3 {(\frac{1}{n})}^{2} (\frac{n - 1}{n}) + \dots$

$∴$ $\frac{n}{9} = (\frac{n - 1}{n}) (1 + 2 (\frac{1}{n}) + 3 {(\frac{1}{n})}^{2} + \dots)$

$\Rightarrow \frac{n}{9} = (\frac{n - 1}{n}) {(1 - \frac{1}{n})}^{- 2} = (\frac{n - 1}{n}) \cdot \frac{n^{2}}{{(n - 1)}^{2}}$

$\Rightarrow \frac{n}{9} = \frac{n}{n - 1} \Rightarrow n = 10$

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