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)

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