Question

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90% ?

Answer 1

Suppose the man tosses a fair coin n times and X denotes the number of heads in n tosses.
$$\begin{gathered} \mathrm{As}\mathit{}p=\frac{1}{2}\text{and}q=\frac{1}{2}, \\ \mathit{}P(X=r)={}^{n}C_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}{\left(\frac{1}{2}\right)}^{n-r},r=0,1,2,3....n \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}P\left(X\ge 1\right)>0.9 \\ \Rightarrow 1-P\left(X=0\right)>0.9 \\ \Rightarrow 1-{}^{n}C_0{\left(\frac{1}{2}\right)}^n>0.9 \\ \Rightarrow {\left(\frac{1}{2}\right)}^n<\frac{1}{10} \\ \Rightarrow 2^n>10 \\ \Rightarrow n=4,5,6.... \\ \mathrm{Hence},\mathrm{the}\mathrm{man}\mathrm{must}\mathrm{toss}\mathrm{the}\mathrm{coin}\mathrm{at}\mathrm{least}4\mathrm{times}. \end{gathered}$$