A coin is tossed n times. The probability that head will turn up an even number of times is

Options

A. `(n+1)/(2n)`
B. `(n)/(n+1)`
C. `1/2`
D. `2^(n-1)`

Answer

To find the probability that a head will turn up an even number of times when a coin is tossed \( n \) times, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- When a coin is tossed \( n \) times, each toss has two outcomes: head (H) or tail (T).
- We need to find the probability of getting an even number of heads.

2. **Identifying the Total Outcomes**:
- The total number of outcomes when tossing a coin \( n \) times is \( 2^n \) since each toss has 2 possible outcomes.

3. **Using Binomial Probability**:
- The probability of getting exactly \( r \) heads in \( n \) tosses is given by the binomial probability formula:
\[ P(X = r) = \binom{n}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{n-r} = \binom{n}{r} \left(\frac{1}{2}\right)^n \]
- Here, \( \binom{n}{r} \) is the binomial coefficient which counts the number of ways to choose \( r \) heads from \( n \) tosses.

4. **Summing Probabilities for Even Heads**:
- We need to sum the probabilities for all even values of \( r \):
\[ P(\text{even heads}) = P(X = 0) + P(X = 2) + P(X = 4) + \ldots \]
- This can be expressed as:
\[ P(\text{even heads}) = \sum_{k=0}^{\lfloor n/2 \rfloor} P(X = 2k) = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \left(\frac{1}{2}\right)^n \]

5. **Utilizing Binomial Theorem**:
- The sum of the coefficients of the even indexed terms in the binomial expansion of \( (x + y)^n \) is given by:
\[ \frac{(1 + 1)^n + (1 - 1)^n}{2} = \frac{2^n + 0}{2} = 2^{n-1} \]
- Therefore, the sum of the probabilities for even heads is:
\[ P(\text{even heads}) = \frac{2^{n-1}}{2^n} = \frac{1}{2} \]

6. **Final Result**:
- Hence, the probability that the head will turn up an even number of times when a coin is tossed \( n \) times is:
\[ \boxed{\frac{1}{2}} \]