A coin is tossed `(2n+1)` times, the probability that head appear odd number of times is

Options

A. `(n)/(2n+1)`
B. `(n+1)/(2n+1)`
C. `1/2`
D. None of these

Answer

To solve the problem of finding the probability that heads appear an odd number of times when a coin is tossed \( (2n + 1) \) times, we can use the properties of binomial distribution.

### Step-by-Step Solution:

1. **Understanding the Problem**:
When a coin is tossed \( (2n + 1) \) times, we can denote the number of heads obtained as \( X \). The random variable \( X \) follows a binomial distribution with parameters \( n = (2n + 1) \) and \( p = \frac{1}{2} \). We need to find the probability that \( X \) is odd.

2. **Using Binomial Probability Formula**:
The probability of getting exactly \( r \) heads in \( n \) tosses is given by the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
Here, \( p = \frac{1}{2} \) and \( 1-p = \frac{1}{2} \).

3. **Calculating the Probability of Odd Heads**:
We want to find the sum of probabilities for all odd values of \( r \):
\[ P(X \text{ is odd}) = P(X = 1) + P(X = 3) + P(X = 5) + \ldots + P(X = 2n + 1) \]
This can be expressed as:
\[ P(X \text{ is odd}) = \sum_{k=0}^{n} P(X = 2k + 1) \]

4. **Substituting Values**:
Each term in the sum can be expressed as:
\[ P(X = 2k + 1) = \binom{2n + 1}{2k + 1} \left(\frac{1}{2}\right)^{2n + 1} \]
Therefore, we have:
\[ P(X \text{ is odd}) = \left(\frac{1}{2}\right)^{2n + 1} \sum_{k=0}^{n} \binom{2n + 1}{2k + 1} \]

5. **Using Binomial Theorem**:
The sum of the binomial coefficients for odd indices can be derived from the binomial expansion:
\[ (1 + 1)^{2n + 1} = 2^{2n + 1} \]
and
\[ (1 - 1)^{2n + 1} = 0 \]
From these, we can deduce:
\[ \sum_{k=0}^{n} \binom{2n + 1}{2k + 1} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} \]

6. **Final Calculation**:
Substituting back into our probability:
\[ P(X \text{ is odd}) = \left(\frac{1}{2}\right)^{2n + 1} \cdot 2^{2n} = \frac{2^{2n}}{2^{2n + 1}} = \frac{1}{2} \]

### Conclusion:
Thus, the probability that heads appear an odd number of times when a coin is tossed \( (2n + 1) \) times is:
\[ \boxed{\frac{1}{2}} \]