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(C) Let $X$ be the number of heads obtained in $7$ tosses. Since the coin is fair,the probability of getting a head is $p = 1/2$ and the probability o

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$1/2$

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(C) Let $X$ be the number of heads obtained in $7$ tosses. Since the coin is fair,the probability of getting a head is $p = 1/2$ and the probability of getting a tail is $q = 1/2$.The person wins if they get more heads than tails. Since there are $7$ tosses,the person wins if they get $4, 5, 6,$ or $7$ heads.The probability is given by the sum of binomial probabilities:$P(X \ge 4) = \sum_{k=4}^{7} \binom{7}{k} (1/2)^k (1/2)^{7-k} = \frac{1}{2^7} \sum_{k=4}^{7} \binom{7}{k}$We know that $\sum_{k=0}^{7} \binom{7}{k} = 2^7 = 128$.Also,by symmetry,$\sum_{k=0}^{3} \binom{7}{k} = \sum_{k=4}^{7} \binom{7}{k} = \frac{128}{2} = 64$.Therefore,the required probability is $\frac{64}{128} = \frac{1}{2}$.

$A$ coin is tossed $7$ times. What is the probability that a person who calls 'heads' every time wins more often?

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