A fair coin is tossed 100 times. The probabil | vedclass

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(A) Let $X$ be the number of tails in $100$ tosses. $X$ follows a binomial distribution with $n = 100$ and $p = \frac{1}{2}$.We need to find the proba

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$\frac{1}{2}$

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(A) Let $X$ be the number of tails in $100$ tosses. $X$ follows a binomial distribution with $n = 100$ and $p = \frac{1}{2}$.We need to find the probability that $X$ is odd, i.e.,$P(X \in \{1, 3, 5, \ldots, 99\})$.The probability is given by $\sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k} p^k q^{n-k}$.Since $p = q = \frac{1}{2}$, this becomes $\sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k} (\frac{1}{2})^{100} = \frac{1}{2^{100}} \sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k}$.We know that the sum of odd-indexed binomial coefficients $\binom{n}{1} + \binom{n}{3} + \ldots = 2^{n-1}$.For $n = 100$, the sum is $2^{100-1} = 2^{99}$.Therefore, the required probability is $\frac{1}{2^{100}} 	imes 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

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

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