1. Define success

A toss is considered a success if the die shows a number less than $n$.

Since the die has faces numbered from $1$ to $n$:

• Numbers less than $n$ are $1,2,3,\dots,n-1$
• So the number of favorable outcomes is $n-1$

Hence, $p = P(\text{number less than } n) = \frac{n-1}{n}$

2. Distribution of number of tosses required

The die is rolled repeatedly until a success occurs, so the number of tosses required follows a geometric distribution with parameter $p$.

For a geometric distribution, the mean is $E(X)=\frac{1}{p}$

Therefore, $E(X)=\frac{1}{(n-1)/n}=\frac{n}{n-1}$

3. Use the given condition

We are given that the mean number of tosses is $\frac{n}{9}$

So, $\frac{n}{n-1} = \frac{n}{9}$

Since $n>1$, we can divide both sides by $n$: $\frac{1}{n-1} = \frac{1}{9}$

Thus, $n-1=9$ $n=10$

4. Final answer

$\boxed{10}$

5. Comparison with stored answer

Stored correct answer: $10$

Our derived answer matches the stored answer.