Question

The probability of a man hitting the target is $\frac{1}{3}$. The number of times must one fire so that the probability of hitting the target atleast once is more than $90\%$ is

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

The correct option is A $6$

Let the number of trials be $n$.
Hence, the probability of the man hitting the target at-least once

$=1-{(1-\frac{1}{3})}^n$
$=1-{\left(\frac{2}{3}\right)}^n$

It is given that the probability should be $90\%=0.9$
Hence,

$1-{\left(\frac{2}{3}\right)}^n=0.9$
$0.1={\left(\frac{2}{3}\right)}^n$

Taking ${\log }_{10}$ on both the sides, we get
$\log \left(0.1\right)=n\left(\log \right(2)-\log (3\left)\right)$
$\Rightarrow -1=n\left(\log \frac{2}{3}\right)$
$\Rightarrow n=\frac{1}{\log \left(3\right)-\log \left(2\right)}$
$\Rightarrow n=5.678$

However, $n$ has to be a natural number.
Hence $n=6$.

Hence, option A.