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If a fair coin is tossed 8 times, then the probability that it shows heads more than tails is
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The correct answer is:
$\frac{93}{256}$
Given $n=8 .$ Here $p=\frac{1}{2}, q=\frac{1}{2}$
$\begin{aligned} P(x>4) &=P(x=5)+P(x=6)+P(x=7)+P(x=8) \\ &={ }^{8} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \\ &=\frac{56}{256}+\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{93}{256} \end{aligned}$
$\begin{aligned} P(x>4) &=P(x=5)+P(x=6)+P(x=7)+P(x=8) \\ &={ }^{8} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \\ &=\frac{56}{256}+\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{93}{256} \end{aligned}$
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