A fair coin is tossed 100 times. The probability of getting head an o…

The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\)

The number of heads X follows the binomial distribution with n = 100 and \(p = \frac{1}{2}\):

    \( \displaystyle P(X = r) = {}^{100}C_{r}\left(\frac{1}{2}\right)^{100} \)

So

    \( \displaystyle P(X \text{ odd}) = \frac{{}^{100}C_{1} + {}^{100}C_{3} + \cdots + {}^{100}C_{99}}{2^{100}} \)

From the binomial expansions of \((1 + 1)^{100}\) and \((1 - 1)^{100}\):

    \(\text{(sum of all } {}^{100}C_{r}) = 2^{100}\)

    \(\text{(even-}r\text{ sum)} - \text{(odd-}r\text{ sum)} = 0\)

So the odd and even sums are equal, each \(2^{99}\):

    \( \displaystyle P(X \text{ odd}) = \frac{2^{99}}{2^{100}} = \frac{1}{2} \)