A fair coin is tossed 100 times . The probability of getting head an odd number of times is

Options

A. `1/2`
B. `1/3`
C. `1/8`
D. `3/8`

Answer

To find the probability of getting heads an odd number of times when a fair coin is tossed 100 times, we can follow these steps:

### Step 1: Define the problem
Let \( n = 100 \) (the number of tosses). We want to find the probability of getting heads an odd number of times.

### Step 2: Identify the random variable
Let \( X \) be the random variable representing the number of heads obtained in 100 tosses. The possible values of \( X \) are \( 0, 1, 2, \ldots, 100 \).

### Step 3: Use the binomial distribution
The number of heads follows a binomial distribution with parameters \( n = 100 \) and \( p = \frac{1}{2} \) (the probability of getting heads in a single toss).

### Step 4: Calculate the probability of odd outcomes
We need to calculate the probability of getting heads an odd number of times, which can be expressed as:
\[ P(X \text{ is odd}) = P(X = 1) + P(X = 3) + P(X = 5) + \ldots + P(X = 99) \]

### Step 5: Use the binomial probability formula
The probability of getting exactly \( k \) heads in \( n \) tosses is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
For our case, \( p = \frac{1}{2} \), so:
\[ P(X = k) = \binom{100}{k} \left(\frac{1}{2}\right)^{100} \]

### Step 6: Factor out the common term
We can factor out \( \left(\frac{1}{2}\right)^{100} \) from the sum:
\[ P(X \text{ is odd}) = \left(\frac{1}{2}\right)^{100} \left( \binom{100}{1} + \binom{100}{3} + \binom{100}{5} + \ldots + \binom{100}{99} \right) \]

### Step 7: Use the binomial theorem
According to the binomial theorem, the sum of the binomial coefficients can be split into odd and even parts:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For odd \( k \):
\[ \sum_{k \text{ odd}} \binom{n}{k} = \frac{1}{2} \cdot 2^n = 2^{n-1} \]
Thus, for \( n = 100 \):
\[ \sum_{k \text{ odd}} \binom{100}{k} = 2^{99} \]

### Step 8: Substitute back into the probability
Now we can substitute this back into our probability expression:
\[ P(X \text{ is odd}) = \left(\frac{1}{2}\right)^{100} \cdot 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2} \]

### Final Answer
The probability of getting heads an odd number of times when a fair coin is tossed 100 times is:
\[ \frac{1}{2} \]

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