Let the probability of hitting the target in one shot be $p=\frac{1}{10}.$ Hence, the probability of missing in one shot is $q=1-p=1-\frac{1}{10}=\frac{9}{10}.$

If the man takes $n$ independent shots, then the probability that he misses all $n$ shots is $\left(\frac{9}{10}\right)^n.$

Therefore, the probability that he hits the target at least once is $1-\left(\frac{9}{10}\right)^n.$

We need this probability to be greater than $\frac14$: $1-\left(\frac{9}{10}\right)^n>\frac14.$

Rearranging, $\left(\frac{9}{10}\right)^n<\frac34.$

Now check the smallest integer $n$:

• For $n=2$, $\left(\frac{9}{10}\right)^2=\frac{81}{100}=0.81>0.75,$ so this does not satisfy the condition.

• For $n=3$, $\left(\frac{9}{10}\right)^3=\frac{729}{1000}=0.729<0.75,$ so this does satisfy the condition.

Hence, the least number of shots required is $n=3.$

Comparison with stored answer:

• Derived answer: $3$
• Stored correct answer: $3$ These agree.