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(C) Let $X$ be the number of heads in $8$ tosses of a fair coin. Here,$n=8$,$p=\frac{1}{2}$,and $q=\frac{1}{2}$.We want to find the probability that h

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$\frac{93}{256}$

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(C) Let $X$ be the number of heads in $8$ tosses of a fair coin. Here,$n=8$,$p=\frac{1}{2}$,and $q=\frac{1}{2}$.We want to find the probability that heads show more than tails,which means $X > 4$.Since the total number of outcomes is $2^8 = 256$,and the distribution is symmetric,we have $P(X  4)$.We know that $\sum_{k=0}^{8} P(X=k) = 1$,so $P(X  4) = 1$.Thus,$2P(X > 4) + P(X=4) = 1$,which implies $P(X > 4) = \frac{1 - P(X=4)}{2}$.Calculating $P(X=4) = {}^{8}C_{4} \left(\frac{1}{2}\right)^{8} = \frac{70}{256}$.Therefore,$P(X > 4) = \frac{1 - \frac{70}{256}}{2} = \frac{\frac{186}{256}}{2} = \frac{93}{256}$.

If a fair coin is tossed $8$ times,then the probability that it shows heads more than tails is

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