A fair coin is tossed 100 times. The chance of getting a head even number of times is

Given $n=100$ tosses of a fair coin with $p=q=\frac{1}{2}$, the number of heads $X$ follows a binomial distribution. The probability of an even number of heads is given by the sum of even binomial coefficients:

$P(\text{even heads}) = \left(\frac{1}{2}\right)^{100} \left[ \binom{100}{0} + \binom{100}{2} + \cdots + \binom{100}{100} \right]$

Using the identity that the sum of even-indexed binomial coefficients equals $2^{n-1}$, we substitute $n=100$ to obtain $2^{99}$.

$P(\text{even heads}) = \frac{1}{2^{100}} \cdot 2^{99} = \frac{1}{2}$

The probability of obtaining an even number of heads is exactly $\frac{1}{2}$.

Answer: C