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

If a fair coin is tossed 8 times, then the probability that it shows heads more than tails is
  1. $\frac{91}{256}$
  2. $\frac{97}{256}$
  3. $\frac{93}{256}$
  4. $\frac{95}{256}$

Solution

Given $n=8 .$ Here $p=\frac{1}{2}, q=\frac{1}{2}$ $\begin{aligned} P(x>4) &=P(x=5)+P(x=6)+P(x=7)+P(x=8) \\ &={ }^{8} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \\ &=\frac{56}{256}+\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{93}{256} \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

Practice more Hyperbola questions on Aicharya