Question

An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.

Answer 1

Let X be the number of heads in tossing the coin 8 times.
X follows a binomial distribution with n =8
$$\begin{gathered} p=\frac{1}{2}\text{and}q=\frac{1}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ \therefore P(X=r)={}^{8}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{8-r},r=0,1,2,3,4,5,6,7,8 \\ \mathrm{Required}\mathrm{probability}=P(X\ge 6) \\ =\mathit{}P(X=6)+P(X=7)+P(X=8) \\ =\frac{{}^{8}C_6+{}^{8}C_7+{}^{8}C_8}{2^8} \\ =\frac{28+8+1}{256} \\ =\frac{37}{256} \end{gathered}$$