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The probability of a man hitting a target is $\frac{1}{10}$. The least number of shots required, so that the probability of his hitting the target at least once is greater than $\frac{1}{4}$, is ______.
$1 - \left(\frac{9}{10}\right)^n > \frac{1}{4} \Rightarrow (0.9)^n < 0.75$. For $n = 3$: $(0.9)^3 = 0.729 < 0.75$. So $n_{\min} = 3$.
BETA
