Question:
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__________.
Solution:
$p=\frac{1}{10}, q=\frac{9}{10}$
$P($ not hitting target in $n$ trials $)=\left(\frac{9}{10}\right)^{n}$
$P($ at least one hit $)=1-\left(\frac{9}{10}\right)^{n}$
$\because 1-\left(\frac{9}{10}\right)^{n}>\frac{1}{4} \Rightarrow(0.9)^{n}<0.75$
$\therefore n_{\text {minimum }}=3$.
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