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 n/9, then n is equal to ____________.

Solution

Maths sir · 4 steps

Each roll is a yes/no trial with success chance (n-1)/n, so the mean wait is just n/(n-1).

1. Success on one roll. A roll succeeds if it shows any of 1, 2, …, n-1, so p = (n-1)/n. It fails only on n, so q = 1/n.
2. Distribution of tosses. Let X be the number of tosses. Needing exactly k tosses means k-1 failures and then a success, so P(X = k) = qᵏ⁻¹p.

   E(X) = Σₖ₌₁^∞ k qᵏ⁻¹p = p/(1-q)² = 1/p = n/(n-1)

3. Match the mean. Set n/(n-1) = n/9. Since n ≠ 0, divide both sides by n to get n - 1 = 9.
4. So. n = 10.

The idea to keep. When you repeat a trial until the first success, the mean number of trials is 1/p. First find p for one trial, and the expected wait follows.

Answer from the NTA final answer key. Solution by PrepWiser.