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

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(A) Let $X$ be the number of tails obtained in $100$ tosses of a fair coin. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$ and $p = \fr

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$\frac{1}{2}$

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(A) Let $X$ be the number of tails obtained in $100$ tosses of a fair coin. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$ and $p = \frac{1}{2}$.The probability of getting tails $k$ times is given by $P(X=k) = {}^{100}C_k (\frac{1}{2})^k (\frac{1}{2})^{100-k} = {}^{100}C_k (\frac{1}{2})^{100}$.We need to find the probability of getting an odd number of tails,which is $P(X=1) + P(X=3) + \dots + P(X=99)$.This sum is equal to $(\frac{1}{2})^{100} 	imes ({}^{100}C_1 + {}^{100}C_3 + \dots + {}^{100}C_{99})$.We know that the sum of odd-indexed binomial coefficients ${}^{n}C_1 + {}^{n}C_3 + \dots = 2^{n-1}$.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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