Question

How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?

Answer 1

Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of X$\ge$1 is more than 80% and X follows a binomial distribution with

$$\begin{gathered} \mathrm{p}=\frac{1}{2},\mathrm{q}=\frac{1}{2} \\ P(X=r)={}^{n}C_r{\left(\frac{1}{2}\right)}^n \\ \text{W}e\mathrm{have}\mathrm{P}(\mathrm{X}\ge 1)=1-\mathrm{P}(\mathrm{X}=0)=1-{}^{n}C_0{\left(\frac{1}{2}\right)}^n=1-\frac{1}{2^{\mathrm{n}}} \\ and\mathrm{P}(\mathrm{X}\ge 1)>80\% \\ 1-\frac{1}{2^{\mathrm{n}}}>80\%=0.80 \\ \frac{1}{2^{\mathrm{n}}}<1-0.80=0.20 \\ {2}^{\mathrm{n}}>\frac{1}{0.2}=5; \\ \mathrm{We}\mathrm{know},2^2<5\mathrm{while}{2}^3>5 \\ \mathrm{So},\mathrm{n}=3 \\ \mathrm{So},n\mathrm{should}\mathrm{be}\mathrm{atleast}3. \end{gathered}$$