An unbiased die is rolled n times. Let P(A), P(B) and P(C ) be the probability of occurrence of an odd number exactly one, two and three times respectively in n trials. If P(A), P(B), P(C ) are in arithmetic progression, then n is equal to

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the problem, we need to find the value of \( n \) such that the probabilities \( P(A) \), \( P(B) \), and \( P(C) \) are in arithmetic progression. Let's break this down step by step.

### Step 1: Define the probabilities
1. **Probability of rolling an odd number**: When rolling a die, the odd numbers are 1, 3, and 5. Therefore, the probability of rolling an odd number is:
\[ P(\text{odd}) = \frac{3}{6} = \frac{1}{2} \]
The probability of rolling an even number is also:
\[ P(\text{even}) = \frac{3}{6} = \frac{1}{2} \]

2. **Probability \( P(A) \)**: This is the probability of getting an odd number exactly once in \( n \) rolls:
\[ P(A) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \cdot \frac{1}{2^n} \]

3. **Probability \( P(B) \)**: This is the probability of getting an odd number exactly twice in \( n \) rolls:
\[ P(B) = \binom{n}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} = \frac{n(n-1)}{2} \cdot \frac{1}{2^n} \]

4. **Probability \( P(C) \)**: This is the probability of getting an odd number exactly three times in \( n \) rolls:
\[ P(C) = \binom{n}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} = \frac{n(n-1)(n-2)}{6} \cdot \frac{1}{2^n} \]

### Step 2: Set up the arithmetic progression condition
According to the problem, \( P(A) \), \( P(B) \), and \( P(C) \) are in arithmetic progression. This means:
\[ 2P(B) = P(A) + P(C) \]

### Step 3: Substitute the probabilities
Substituting the expressions we derived:
\[ 2 \cdot \frac{n(n-1)}{2} \cdot \frac{1}{2^n} = n \cdot \frac{1}{2^n} + \frac{n(n-1)(n-2)}{6} \cdot \frac{1}{2^n} \]

### Step 4: Simplify the equation
We can cancel \( \frac{1}{2^n} \) from both sides (assuming \( n \geq 1 \)):
\[ n(n-1) = n + \frac{n(n-1)(n-2)}{6} \]

### Step 5: Multiply through by 6 to eliminate the fraction
\[ 6n(n-1) = 6n + n(n-1)(n-2) \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ 6n^2 - 6n = 6n + n^3 - 3n^2 + 2n \]
\[ 0 = n^3 - 3n^2 + 2n + 6n - 6n^2 \]
\[ 0 = n^3 - 9n^2 + 8n \]

### Step 7: Factor the polynomial
Factoring out \( n \):
\[ n(n^2 - 9n + 8) = 0 \]
This gives:
\[ n = 0 \quad \text{or} \quad n^2 - 9n + 8 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ n = \frac{9 \pm \sqrt{(9)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ n = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2} \]
This gives:
\[ n = \frac{16}{2} = 8 \quad \text{or} \quad n = \frac{2}{2} = 1 \]

### Step 9: Choose the valid solution
Since we need \( n \) to be at least 3 for \( P(C) \) to be defined, we choose \( n = 8 \).

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{8} \]