To solve the problem step by step, we will calculate the probabilities P(A), P(B), and P(C) based on the rolling of an unbiased die and then determine the value of n such that these probabilities are in arithmetic progression.
### Step 1: Determine the probability of rolling an odd number
When a die is rolled, the total outcomes are 6 (1, 2, 3, 4, 5, 6). The favorable outcomes for rolling an odd number are 3 (1, 3, 5). Thus, the probability of rolling an odd number (p) is:
\[
p = \frac{3}{6} = \frac{1}{2}
\]
### Step 2: Calculate P(A), P(B), and P(C)
- **P(A)**: Probability of rolling an odd number exactly once in n trials:
\[
P(A) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = \binom{n}{1} \left(\frac{1}{2}\right)^n = n \left(\frac{1}{2}\right)^n
\]
- **P(B)**: Probability of rolling an odd number exactly twice in n trials:
\[
P(B) = \binom{n}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} = \binom{n}{2} \left(\frac{1}{2}\right)^n = \frac{n(n-1)}{2} \left(\frac{1}{2}\right)^n
\]
- **P(C)**: Probability of rolling an odd number exactly three times in n trials:
\[
P(C) = \binom{n}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} = \binom{n}{3} \left(\frac{1}{2}\right)^n = \frac{n(n-1)(n-2)}{6} \left(\frac{1}{2}\right)^n
\]
### Step 3: Set up the arithmetic progression condition
Given that P(A), P(B), and P(C) are in arithmetic progression, we have:
\[
2P(B) = P(A) + P(C)
\]
Substituting the expressions we found:
\[
2 \cdot \frac{n(n-1)}{2} \left(\frac{1}{2}\right)^n = n \left(\frac{1}{2}\right)^n + \frac{n(n-1)(n-2)}{6} \left(\frac{1}{2}\right)^n
\]
Cancelling \(\left(\frac{1}{2}\right)^n\) from both sides (since it is non-zero):
\[
n(n-1) = n + \frac{n(n-1)(n-2)}{6}
\]
### Step 4: Simplifying the equation
Multiply through by 6 to eliminate the fraction:
\[
6n(n-1) = 6n + n(n-1)(n-2)
\]
Expanding both sides:
\[
6n^2 - 6n = 6n + n^3 - 3n^2 + 2n
\]
Rearranging gives:
\[
n^3 - 3n^2 + 2n - 6n - 6n^2 = 0
\]
\[
n^3 - 9n^2 + 2n = 0
\]
Factoring out n:
\[
n(n^2 - 9n + 2) = 0
\]
### Step 5: Solve the quadratic equation
The quadratic equation is:
\[
n^2 - 9n + 2 = 0
\]
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
n = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 8}}{2} = \frac{9 \pm \sqrt{73}}{2}
\]
Calculating the roots gives two values, but we also have \(n = 0\) from the factorization.
### Step 6: Determine valid n
Since n must be a positive integer and at least 3 (as per the problem statement), we check the roots:
1. \(n = 7\) (valid)
2. \(n = 2\) (not valid)
Thus, the only valid solution is:
\[
\boxed{7}
\]
