Question

A coin is tossed 5 times. What is the probability of getting at least 3 heads?

Answer 1

$$\begin{gathered} \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}\mathrm{in}5\mathrm{tosses}. \\ X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n\mathit{}=5;p=\mathrm{prob}\text{ability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{head}=\frac{1}{2}\mathrm{and}q=1-p=\frac{1}{2} \\ \mathrm{P}(X=r)={}^{5}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{5-r},r=0,1,2...5 \\ \mathrm{The}\mathrm{required}\mathrm{probability}=P(\mathrm{getting}\mathrm{at}\mathrm{least}3\mathrm{heads}) \\ =P(X\ge 3) \\ =\mathit{}P(X=3)+P(X=4)+P(X=5) \\ ={}^{5}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{5-3}+{}^{5}C_4{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^1{\left(\frac{1}{2}\right)}^{5-1}+{}^{5}C_5{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^5{\left(\frac{1}{2}\right)}^{5-0} \end{gathered}$$
$$\begin{gathered} =10{\left(\frac{1}{2}\right)}^5+5{\left(\frac{1}{2}\right)}^5+1{\left(\frac{1}{2}\right)}^5 \\ ={\left(\frac{1}{2}\right)}^5(10+5+1) \\ ={\left(\frac{1}{2}\right)}^5\times 16 \\ =\frac{1}{2} \end{gathered}$$