Question

Six coins are tossed simultaneously. Find the probability of getting
(i) 3 heads
(ii) no heads
(iii) at least one head

Answer 1

Let X denote the number of heads obtained in tossing 6 coins.
Then, X follows a binomial distribution with n=6,

$$\begin{gathered} p=\frac{1}{2}\text{and}q=\frac{1}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{6}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{6-r},\mathit{}r=0,1,2,3,4,5,6 \\ =\frac{{}^{6}C_r}{2^6} \\ \left(\mathrm{i}\right)P(\mathrm{getting}3\hspace{0.17em}\mathrm{heads})=P(X=3) \\ =\frac{{}^{6}C_3}{2^6} \\ =\frac{20}{64} \\ =\frac{5}{16} \\ \left(\mathrm{ii}\right)P(\mathrm{getting}\mathrm{no}\mathrm{head})=P(\mathrm{X}=0) \\ =\frac{{}^{6}C_0}{2^6} \\ ={\left(\frac{1}{2}\right)}^6 \\ =\frac{1}{64} \\ \left(\mathrm{iii}\right)P(\mathrm{getting}\mathrm{at}\mathrm{least}1\mathrm{head})=\mathit{}P(X\ge 1) \\ =1-\mathit{}P(X=0) \\ =1-\frac{1}{64} \\ =\frac{63}{64} \end{gathered}$$