Question

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
[NCERT EXEMPLAR]

Answer 1

Let hitting the target be a success in a shoot.

We have,

$$\begin{gathered} p=\mathrm{probability}\mathrm{of}\mathrm{hitting}\mathrm{the}\mathrm{target}=0.25=\frac{1}{4} \\ \mathrm{Also},q=1-p=1-\frac{1}{4}=\frac{3}{4} \\ \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{success}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}7\mathrm{trials}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=7\mathrm{and}p=\frac{1}{4} \\ \therefore \mathrm{P}\left(X=r\right)={}^7\mathrm{C}_r{p}^r{q}^{\left(7-r\right)}={}^7\mathrm{C}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{\left(7-r\right)}=\frac{{}^7\mathrm{C}_r{3}^{\left(7-r\right)}}{4^7},\mathrm{where}r=0,1,2,3,4,5 \\ \mathrm{Now}, \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(X\ge 2\right) \\ =1-\left[\mathrm{P}\left(X=0\right)+\mathrm{P}\left(X=1\right)\right] \\ =1-\left[\frac{{}^7\mathrm{C}_0{3}^7}{4^7}+\frac{{}^7\mathrm{C}_1{3}^6}{4^7}\right] \\ =1-\left[\frac{2187}{16384}+\frac{5103}{16384}\right] \\ =1-\frac{7290}{16384} \\ =\frac{9094}{16384} \\ =\frac{4547}{8192} \end{gathered}$$