Question

If the probability of hitting a target by a shooter, in any shot, is $\frac{1}{3}$, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than $\frac{5}{6}$, is :

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

Let the number of independent shots at the target required by him $=n$
Probility of not hitting the target in an attempt is $\frac{2}{3}$
Therefore, probability of not hitting the target in all $n$ attempts is ${\left(\frac{2}{3}\right)}^n$
$\Rightarrow 1-{\left(\frac{2}{3}\right)}^n>\frac{5}{6}$
$\Rightarrow {\left(\frac{2}{3}\right)}^n<\frac{1}{6}$
$\Rightarrow \left(\frac{2^{n+1}}{3^{n-1}}\right)<1$
$\Rightarrow 3^{n-1}>2^{n+1}$

Hence, $n_{min}=5$