Question

A coin is tossed 5 times. What is the probability that tail appears an odd number of times?

Answer 1

$$\begin{gathered} \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tails}\mathrm{when}\text{a}\mathrm{coin}\mathrm{is}\mathrm{tossed}5\mathrm{times}. \\ X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n\mathit{}=5;p=\frac{1}{2};q=1-p=\frac{1}{2} \\ \mathrm{Then}P(X=r)={}^{5}C_r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^{\mathrm{r}}{\left(\frac{1}{2}\right)}^{n-r}={}^{5}C_r{\left(\frac{1}{2}\right)}^5 \\ \mathrm{The}\mathrm{required}\mathrm{probability}=P(X=\mathrm{odd}) \\ =P(X=1)+P(X=3)+P(X=5) \\ ={}^{5}C_1{\left(\frac{1}{2}\right)}^5+{}^{5}C_3{\left(\frac{1}{2}\right)}^5+{}^{5}C_5{\left(\frac{1}{2}\right)}^5 \\ ={\left(\frac{1}{2}\right)}^5[5+10+1] \\ =\frac{16}{32} \\ =\frac{1}{2} \end{gathered}$$