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

The given mean is expressed as:

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

Simplify the series:

\( \frac{n}{9} = \frac{n-1}{n} \left( 1 + 2 \cdot \frac{1}{n} + 3 \cdot \left( \frac{1}{n} \right)^2 + \dots \right) \)

The infinite series inside the parentheses is a geometric series:

\( 1 + 2 \cdot \frac{1}{n} + 3 \cdot \left( \frac{1}{n} \right)^2 + \dots \)

Using the sum formula for such series:

\( \frac{n}{9} = \frac{n-1}{n} \cdot \left( 1 - \frac{1}{n} \right)^{-2} \)

Simplify further:

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

Multiply through:

\( \frac{n}{9} = \frac{n}{n-1} \)

Solve for \( n \):

\( n - 1 = 9 \Rightarrow n = 10 \)