Question

A fair n-faced die is rolled until a number less than n appears. If the mean of tosses is $n/9$, then n =

• A. 4
• B. 6
• C. 8
• D. 10

Hint:

Mean tosses until success is always the reciprocal of the success probability.

Answer 1

The Correct Option is D

Step 1: Concept
This follows a Geometric Distribution where $p$ is the probability of success (number $< n$).

Step 2: Analysis
On an $n$-faced die, numbers are $1, 2, ..., n$. Success is $\{1, 2, ..., n-1\}$.
$p = \frac{n-1}{n}$.

Step 3: Calculation
Mean of Geometric Distribution is $1/p = \frac{n}{n-1}$.
Given Mean $= n/9 \implies \frac{n}{n-1} = \frac{n}{9}$.
So $n-1 = 9 \implies n = 10$.

Step 4: Conclusion
Hence, $n = 10$. Final Answer: (D)