Question

Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws.
[NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{getting}\mathrm{a}\mathrm{multiple}\mathrm{of}3\mathrm{be}\mathrm{a}\mathrm{success}. \\ \mathrm{We}\mathrm{have}, \\ p=\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{success}=\frac{2}{6}=\frac{1}{3} \\ \mathrm{So},q=\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{failure}=1-p=1-\frac{1}{3}=\frac{2}{3} \\ \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{success}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}10\mathrm{trials}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=10,p=\frac{1}{3}\mathrm{and}q=\frac{2}{3} \\ \therefore \mathrm{P}\left(X=r\right)={}^{10}\mathrm{C}_r{p}^r{q}^{\left(10-r\right)}={}^{10}\mathrm{C}_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{\left(10-r\right)}={}^{10}\mathrm{C}_r{\left(\frac{1}{3}\right)}^{10}{2}^{\left(10-r\right)}=\frac{{}^{10}\mathrm{C}_r{2}^{\left(10-r\right)}}{3^{10}},r=0,1,2,...,10 \\ \mathrm{Now}, \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(X\ge 8\right) \\ =\mathrm{P}\left(X=8\right)+\mathrm{P}\left(X=9\right)+\mathrm{P}\left(X=10\right) \\ =\frac{{}^{10}\mathrm{C}_8{2}^{\left(10-8\right)}}{3^{10}}+\frac{{}^{10}\mathrm{C}_9{2}^{\left(10-9\right)}}{3^{10}}+\frac{{}^{10}\mathrm{C}_{10}{2}^{\left(10-10\right)}}{3^{10}} \\ =\frac{45\times 2^2}{3^{10}}+\frac{10\times 2}{3^{10}}+\frac{1}{3^{10}} \\ =\frac{180+20+1}{3^{10}} \\ =\frac{201}{3^{10}} \end{gathered}$$