The probability of a man hitting a target is $$\frac{1}{4}$$. The number of times he must shoot so that the probability he hits the target, at least once is more than $$0.9,$$  is :
$$\left[ {{\text{use }}\log \,4 = 0.602{\text{ and }}\log \,3 = 0.477} \right]$$

Solution :
Let $$n$$ denote the required number of shots and $$X$$ the number of shots that hit the target.
Then $$X \sim B\left( {n,\,p} \right),$$    with $$p = \frac{1}{4}.$$
Now,
$$\eqalign{ & P\left( {X \geqslant 1} \right) \geqslant 0.9 \cr & \Rightarrow 1 - P\left( {X = 0} \right) \geqslant 0.9 \cr & \Rightarrow 1 - {}^n{C_0}{\left( {\frac{3}{4}} \right)^n} \geqslant 0.9 \cr & \Rightarrow {\left( {\frac{3}{4}} \right)^n} \leqslant \frac{1}{{10}} \cr & \Rightarrow {\left( {\frac{4}{3}} \right)^n} \geqslant 10 \cr & \Rightarrow n\left( {\log \,4 - \log \,3} \right) \geqslant 1 \cr & \Rightarrow n\left( {0.602 - 0.477} \right) \geqslant 1 \cr & \Rightarrow n \geqslant \frac{1}{{0.125}} = 8 \cr} $$
Therefore the least number of trials required is $$8.$$