The probability of a man hitting a target is 2/5. He first at the target k times (k, a given number). Then the minimum k, so that the probability of hitting the target at least once is more than 7/10 is

Given Probability of a man hitting target is $\frac{2}{5}$

and Probability of a man not hitting the target is $1-\frac{2}{5}=\frac{3}{5}$

Probability of hitting the target in one fire $=\frac{2}{5}$

Probability of the hitting the target in two fires $=\frac{3}{5}\times \frac{2}{5}$

Probability of hitting the target in three fires $\frac{3}{5}\times \frac{3}{5}\times \frac{2}{5}$ and similarly it goes up to k times

Now Probability of hitting the target at least 

Once $=\frac{2}{5}+\frac{3}{5}\times \frac{2}{5}+\frac{3}{5}\times \frac{3}{5}\times \frac{2}{5}+.......$k times

          $$\begin{gathered} =\frac{2}{5} \left[1+\frac{3}{5}+\frac{3}{5}\times \frac{3}{5}+.......+\mathrm{k} \mathrm{times}\right] \\ =\frac{2}{5} \left[\frac{1-{\left({\displaystyle \frac{3}{5}}\right)}^{\mathrm{k}}}{1-{\displaystyle \frac{3}{5}}}\right] \\ =\frac{2}{5}\left[\frac{1-{\left({\displaystyle \frac{3}{5}}\right)}^{\mathrm{k}}}{{\displaystyle \frac{2}{5}}}\right] \\ =1-{\left(\frac{3}{5}\right)}^{\mathrm{k}} \end{gathered}$$

Given that $1-{\left(\frac{3}{5}\right)}^{\mathrm{k}}>\frac{7}{10}$

          $$\begin{gathered} \Rightarrow \frac{3}{10}>{\left(\frac{3}{5}\right)}^{\mathrm{k}} \\ \Rightarrow {\left(\frac{3}{5}\right)}^{\mathrm{k}}<0.3 \\ \Rightarrow {(0.6)}^{\mathrm{k}}<0.3 \\ \Rightarrow \mathrm{k} \log (0.6)>\log (0.3) \\ \Rightarrow \mathrm{k}>\frac{\log (0.3)}{\log (0.6)} \\ \Rightarrow \mathrm{k}>2.36 \end{gathered}$$

$\therefore$ minimum Value of k=3