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(D) Let $X$ be the random variable representing the number of tosses required to get a number less than $n$. This follows a geometric distribution whe

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$10$

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(D) Let $X$ be the random variable representing the number of tosses required to get a number less than $n$. This follows a geometric distribution where the probability of success $p$ is the probability of getting a number from ${1, 2, \dots, n-1}$. Thus,$p = \frac{n-1}{n}$. The mean of a geometric distribution is given by $E[X] = \frac{1}{p}$. Given $E[X] = \frac{n}{9}$,we have $\frac{1}{p} = \frac{n}{9}$. Substituting $p = \frac{n-1}{n}$,we get $\frac{n}{n-1} = \frac{n}{9}$. Since $n \in N$ and $n > 1$,we can divide by $n$ to get $\frac{1}{n-1} = \frac{1}{9}$. Therefore,$n-1 = 9$,which implies $n = 10$.

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

$A$ fair $n$-faced 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 $n=$ (where $n \in N$ ).

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