A player tosses a coin and score one point for every head and two points for every tail that turns up. He plays on until his score reaches or passes n. `P_(n)` denotes the probability of getting a score of exactly n.
The value of `P_(n)+(1//2)P_(n-1)` is equal to

Options

A. (a) `1//2`
B. (b) `2//3`
C. (c) 1
D. (d) None of these

Explanation

To solve the problem, we need to analyze the scoring system and the probabilities involved when a player tosses a coin. The player scores 1 point for every head and 2 points for every tail. We denote \( P(n) \) as the probability of obtaining a score of exactly \( n \).

### Step-by-step Solution:

1. **Understanding the Scoring System**:
- Each head contributes 1 point.
- Each tail contributes 2 points.
- The player continues tossing the coin until the score reaches or exceeds \( n \).

2. **Setting Up the Probabilities**:
- Let \( P(n) \) be the probability of scoring exactly \( n \).
- The player can reach a score of \( n \) in two ways:
- By tossing a head (which gives \( n-1 \) points) and then getting a head.
- By tossing a tail (which gives \( n-2 \) points) and then getting a tail.

3. **Formulating the Probability**:
- The probability of scoring \( n \) can be expressed as:
\[ P(n) = \frac{1}{2} P(n-1) + \frac{1}{2} P(n-2) \]
- Here, \( \frac{1}{2} \) is the probability of getting either a head or a tail.

4. **Finding \( P(n) + \frac{1}{2} P(n-1) \)**:
- We need to find the value of \( P(n) + \frac{1}{2} P(n-1) \).
- Substituting the expression for \( P(n) \):
\[ P(n) + \frac{1}{2} P(n-1) = \left(\frac{1}{2} P(n-1) + \frac{1}{2} P(n-2)\right) + \frac{1}{2} P(n-1) \]
- This simplifies to:
\[ = \frac{1}{2} P(n-2) + \frac{1}{2} P(n-1) + \frac{1}{2} P(n-1) \]
- Combining the terms gives:
\[ = \frac{1}{2} P(n-2) + P(n-1) \]

5. **Final Calculation**:
- Since we have established that \( P(n) + \frac{1}{2} P(n-1) = P(n-1) + \frac{1}{2} P(n-2) \), we can analyze the probabilities further.
- The recurrence relation indicates that the total probability will converge to a specific value.
- After evaluating the probabilities and their contributions, we find that:
\[ P(n) + \frac{1}{2} P(n-1) = 1 \]
- Thus, the final answer is:
\[ \boxed{1} \]