Question

A coin is tossed n times. The probability of getting at least once is greater than 0.8. Then, the least value of n, is
(a) 2
(b) 3
(c) 4
(d) 5

Answer 1

(b) 3

Let X be the number of heads. Then X follows a binomial distribution with

$$\begin{gathered} p=\frac{1}{2},\mathit{}q=\frac{1}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)=\mathit{}{}^{n}C_r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^{n\mathit{-}r},r=0,1,2,3..\mathit{.}\mathit{.}\mathit{.}\mathit{.}n \\ \therefore P(X\ge 1)=1-P(X=0) \\ =1-{\left(\frac{1}{2}\right)}^n\ge 0.8 \\ \text{Or}{2}^n\ge \frac{1}{0.2} \\ \Rightarrow 2^n\ge 5 \\ \mathrm{This}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}n\ge 3. \\ \mathrm{So},\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}n\mathit{}\mathrm{must}\mathrm{be}3. \end{gathered}$$