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)` is equal to

Options

A. (a) `(1//2)[P_(n-1)+P_(n-2)]`
B. (b) `(1//2)[2P_(n-1)+P_(n-2)]`
C. (c) `(1//2)[P_(n-1)+2P_(n-2)]`
D. (d) None of these

Explanation

To solve the problem, we need to determine the probability \( P(n) \) of scoring exactly \( n \) points when a player tosses a coin, scoring 1 point for each head and 2 points for each tail.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The player scores 1 point for a head (H) and 2 points for a tail (T).
- The player continues tossing the coin until the score reaches or exceeds \( n \).

2. **Defining the Probability**:
- Let \( P(n) \) be the probability of scoring exactly \( n \).
- The player can reach a score of \( n \) by either:
- Tossing a head last (which means the previous score was \( n-1 \)).
- Tossing a tail last (which means the previous score was \( n-2 \)).

3. **Setting Up the Equation**:
- The probability of getting a head is \( \frac{1}{2} \) and the probability of getting a tail is \( \frac{1}{2} \).
- Therefore, we can express \( P(n) \) as:
\[ P(n) = \frac{1}{2} P(n-1) + \frac{1}{2} P(n-2) \]

4. **Base Cases**:
- We need to establish base cases for \( P(n) \):
- \( P(0) = 1 \): The probability of scoring 0 points (not tossing the coin at all).
- \( P(1) = \frac{1}{2} \): The player can only score 1 point by tossing one head.
- \( P(2) = \frac{1}{2} P(1) + \frac{1}{2} P(0) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \).

5. **Recursive Calculation**:
- Using the recursive relation, we can calculate \( P(n) \) for any \( n \) based on the previous two probabilities:
\[ P(n) = \frac{1}{2} P(n-1) + \frac{1}{2} P(n-2) \]

6. **Final Expression**:
- The final expression for \( P(n) \) can be computed recursively or iteratively using the base cases and the established relation.