The probability of a man hitting a target is $\frac{1}{10}$. The least number of shots required, so

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

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(3) : Here, $p=\frac{1}{10}, q=\frac{9}{10}$\ $	herefore \quad P$ (not hitting in $n$ trials) $=\left(\frac{9}{10}ight)^{n}$\ $\Rightarrow \quad P$ (hitting the target atleast once)\ $=1-\left(\frac{9}{10}ight)^{n}>\frac{1}{4}$\ $\Rightarrow \quad\left(\frac{9}{10}ight)^{n}<\frac{3}{4}$, which is true for $n=3$\ So, least number of shots required is 3 .\