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

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(A) Let $X$ be the number of heads in $n=100$ tosses. Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probabilit

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$\frac{1}{2}$

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(A) Let $X$ be the number of heads in $n=100$ tosses. Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.$X$ follows a binomial distribution $B(n, p) = B(100, \frac{1}{2})$.The probability of getting an even number of heads is $P(X \in \{0, 2, 4, \dots, 100\}) = \sum_{k \in \{0, 2, \dots, 100\}} \binom{100}{k} p^k q^{100-k}$.Since $p=q=\frac{1}{2}$,this becomes $\left(\frac{1}{2}ight)^{100} \sum_{k \in \{0, 2, \dots, 100\}} \binom{100}{k}$.We know that the sum of even binomial coefficients is $\sum_{k 	ext{ even}} \binom{n}{k} = 2^{n-1}$.Thus,the probability is $\left(\frac{1}{2}ight)^{100} 	imes 2^{100-1} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

$A$ fair coin is tossed $100$ times. The probability of getting a head an even number of times is

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