Question

A fair coin is tossed 6 times. What is the probability of getting at least 3 heads?

Answer 1

$$\begin{gathered} \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}\mathrm{when}6\mathrm{coins}\mathrm{are}\mathrm{tossed}. \\ \mathrm{Then},X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n\mathit{}=6;\mathit{}p\mathit{}=\frac{1}{2}\mathrm{and}q=\frac{1}{2}; \\ P(X=r)={}^{6}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{6-r} \\ ={}^{6}C_r{\left(\frac{1}{2}\right)}^5,\mathit{}r=0,1,2.....6 \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{required}\mathrm{probability}=\mathit{}P(X\ge 3) \\ =P(X=3)+P(X=4)+P(X=5)+P(X=6) \\ ={}^{6}C_3{\left(\frac{1}{2}\right)}^6+{}^{6}C_4{\left(\frac{1}{2}\right)}^6+{}^{6}C_5{\left(\frac{1}{2}\right)}^6+{}^{6}C_6{\left(\frac{1}{2}\right)}^6 \\ =(20+15++6+1)\frac{1}{64} \\ =\frac{21}{32} \end{gathered}$$