Question

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_____.

Hint:

For problems involving expected values, carefully consider the probabilities of each outcome and use the formula for expectation: E(X) = xP(X = x)

Answer 1

Correct Answer: 10

To determine the value of \( n \), we need to analyze the problem of rolling an n-faced die until a number less than \( n \) appears. Since the die has faces numbered \( 1, 2, \ldots, n \), we can conclude that any number less than \( n \) appearing involves numbers \( 1 \) to \( n-1 \).
The probability of rolling a number less than \( n \) in a single toss is \(\frac{n-1}{n}\) since there are \( n-1 \) favorable outcomes.
Thus, the probability of rolling the number \( n \) is \(\frac{1}{n}\).
The expected number of rolls \( E \) required to get a number less than \( n \) follows a geometric distribution with success probability \( \frac{n-1}{n} \). The mean (expected value) \( E \) of a geometric distribution with success probability \( p \) is given by \( E=\frac{1}{p} \).
In this context, \( p = \frac{n-1}{n} \), so:
\( E = \frac{1}{\frac{n-1}{n}} = \frac{n}{n-1} \). 
We know from the problem statement that \( E = \frac{n}{9} \).
Thus, we have the equation:
\( \frac{n}{n-1} = \frac{n}{9} \).
Cross-multiplying gives:
\( 9n = n(n-1) \).
Expanding and simplifying leads to:
\( 9n = n^2 - n \) or \( n^2 - 10n = 0 \).
Factoring out \( n \) yields:
\( n(n-10) = 0 \).
Since \( n > 1 \), we discard \( n=0 \), and thus \( n = 10 \).
Verifying the solution, \( n = 10 \) fits the given range of 10,10. Therefore, the value of \( n \) is \( \boxed{10} \).