A player tosses a coin. He sets one point for head and 2 points for tail. He plays till he gets sum of points equal to n. If `p_(n)` be the probability that his score becomes n, then

Options

A. `p_(3)=(1)/(2)`
B. `p_(n)=(1)/(2)p_(n-1)+(1)/(4)p_(n-2)`.
C. `p_(4)=(11)/(16)`
D. `p_(n)=(1)/(2)(p_(n-1)+p_(n-2))`.

Answer

To solve the problem of finding the probability \( p_n \) that the player's score becomes \( n \) when tossing a coin where heads score 1 point and tails score 2 points, we can break it down step by step.

### Step 1: Understanding the Problem
The player can score points by tossing a coin:
- Heads (H) gives 1 point.
- Tails (T) gives 2 points.

The player continues tossing until the total score equals \( n \).

### Step 2: Finding Possible Combinations
To achieve a score of \( n \), we need to consider different combinations of heads and tails:
1. If the player gets \( k \) tails, then the score from tails is \( 2k \).
2. The remaining score must come from heads, which means the total score from heads must be \( n - 2k \).

Let \( h \) be the number of heads. Then, we have:
\[ h + 2k = n \]
From this, we can express \( h \) as:
\[ h = n - 2k \]

### Step 3: Valid Combinations
For the combination to be valid:
- \( h \) must be non-negative, which implies \( n - 2k \geq 0 \) or \( k \leq \frac{n}{2} \).
- The number of tosses is \( h + k = (n - 2k) + k = n - k \).

### Step 4: Counting Arrangements
The total number of ways to arrange \( h \) heads and \( k \) tails in \( n - k \) tosses is given by the binomial coefficient:
\[ \binom{n - k}{k} \]
This counts the ways to choose \( k \) positions for tails in \( n - k \) tosses.

### Step 5: Probability Calculation
The probability \( p_n \) that the score becomes \( n \) can be calculated as:
\[ p_n = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k} \left( \frac{1}{2} \right)^{n-k} \left( \frac{1}{2} \right)^{k} \]
This simplifies to:
\[ p_n = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k} \left( \frac{1}{2} \right)^{n} \]

### Step 6: Final Expression
Thus, we can express \( p_n \) as:
\[ p_n = \frac{1}{2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k} \]

### Summary
The probability \( p_n \) that the player's score becomes \( n \) is given by:
\[ p_n = \frac{1}{2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k} \]