Question

A can hit a target 3 times in 6 shots, B : 2 times in 6 shots and C : 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

Answer 1

$$\begin{gathered} P\left(\text{A hits the target}\right)=\frac{3}{6} \\ P\left(\text{B hits the target}\right)=\frac{2}{6} \\ P\left(\text{C hits the target}\right)=\frac{4}{4}=1 \\ P\left(\text{atleast 2 shots hit}\right)=P\left(\text{exactly 2 shots hit}\right)+P\left(\text{all 3 shots hit}\right) \\ =\frac{3}{6}\left(1-\frac{2}{6}\right)+\frac{2}{6}\left(1-\frac{3}{6}\right)+\frac{3}{6}\times \frac{2}{6}\times 1\left(\mathrm{Here},\mathrm{the}\mathrm{probability}\mathrm{of}C\mathrm{hitting}\mathrm{the}\mathrm{target}\mathrm{is}1.\mathrm{S}o,\mathrm{it}\mathrm{will}\mathrm{always}\mathrm{hit}. \\ \mathrm{When}\mathrm{exactly}2\mathrm{shots}\mathrm{are}\mathrm{hit},\mathrm{then}\mathrm{either}A\mathrm{hits}\mathrm{or}B\mathrm{hits}.\right) \\ =\frac{3}{6}\times \frac{4}{6}+\frac{2}{6}\times \frac{3}{6}+\frac{6}{36} \\ =\frac{12+6+6}{36} \\ =\frac{24}{36} \\ =\frac{2}{3} \end{gathered}$$