A man firing at a distant target has $$10\% $$  chance of hitting the target in one shot. The number of times he must fire at the target to have about $$50\% $$  chance of hitting the target is :

Solution :
The probability of hitting in one shot $$ = \frac{{10}}{{100}} = \frac{1}{{10}}$$
 If he fires $$n$$ shots, the probability of hitting at least once
$$\eqalign{ & = 1 - {\left( {1 - \frac{1}{{10}}} \right)^n} = 1 - {\left( {\frac{9}{{10}}} \right)^n} = \frac{1}{2}\,\,\left( {{\text{from the question}}} \right) \cr & \therefore \,{\left( {\frac{9}{{10}}} \right)^n} = \frac{1}{2} \cr & \therefore \,n\left\{ {2{{\log }_{10}}3 - 1} \right\} = - {\log _{10}}2 \cr & \therefore \,n = \frac{{{{\log }_{10}}2}}{{1 - 2{{\log }_{10}}3}} = \frac{{0.3010}}{{1 - 2 \times 0.4771}} = 6.5\left( {{\text{nearly}}} \right) \cr} $$
$$\therefore $$  for $$6$$ shots, the probability is about $$53\% $$  while for $$7$$ shots it is nearly $$48\% .$$