The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times $(k$, a given number). Then the minimum $k$, so that the probability of hitting the target at least once is more than $\frac{7}{10}$, is :

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Aicharya · Accepted Answer

$ \begin{aligned} & \frac{2}{5}+\frac{3}{5} 	imes \frac{2}{5}+\left(\frac{3}{5}ight)^2 	imes \frac{2}{5}+\ldots \ldots+\left(\frac{3}{5}ight)^k \cdot \frac{2}{5}>\frac{7}{10} \ \Rightarrow & \frac{2}{5}\left[1+\frac{3}{5}+\left(\frac{3}{5}ight)^2+\ldots \ldots+\left(\frac{3}{5}ight)^kight]>\frac{7}{10} \ \Rightarrow & \frac{2}{5} 	imes \frac{1-\left(\frac{3}{5}ight)^k}{1-\frac{3}{5}}>\frac{7}{10} \Rightarrow 1-\left(\frac{3}{5}ight)^k>\frac{7}{10} \ \Rightarrow & \left(\frac{3}{5}ight)^k < \frac{3}{10} \Rightarrow k \geq 3 \end{aligned} $ Hence minimum value of $k=3$

The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times $(k$, a given…

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