Question

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 :

• A. 3
• B. 5
• C. 2
• D. 4

Answer 1

Answer: (A)

$\frac{2}{5}+\frac{3}{5}\times\frac{2}{5}+\left(\frac{3}{5}\right)^{2}\times\frac{2}{5}+\ldots\ldots+\left(\frac{3}{5}\right)^{k}.\frac{2}{5} > \frac{7}{10}$
$\Rightarrow \quad \frac{2}{5}\left[1+\frac{3}{5}+\left(\frac{3}{5}\right)^{2}+\ldots\ldots +\left(\frac{3}{5}\right)^{k}\right] > \frac{7}{10}$
$\Rightarrow \quad \frac{5}{5}\times \frac{1-\left(\frac{3}{5}\right)^{k}}{1-\frac{3}{5}} > \frac{7}{10}\quad\Rightarrow \quad1-\left(\frac{3}{5}\right)^{k}> \frac{7}{10}$
$\Rightarrow \quad \left(\frac{3}{5}\right)^{k}> \frac{3}{10}\quad\Rightarrow \quad k \ge 3$
Hence minimum value of $k = 3$