Question

The probability of a shooter hitting a target is \[\frac{3}{4} .\] How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

 

Answer 1

Let the shooter fire n times and let X denote the number of times the shooter hits the target.
Then, X follows binomial distribution with \[p = \frac{3}{4}\text{ and }  q = \frac{1}{4}\]  such that 

\[P\left( X = r \right) = ^{n}{}{C}_r \left( \frac{3}{4} \right)^r \left( \frac{1}{4} \right)^{n - r} \]
\[ \Rightarrow P\left( X = r \right) = ^{n}{}{C}_r \frac{3^r}{4^n}\]
\[\text{ It is given that } P\left( X \geq 1 \right) > 0 . 99\]
\[ \Rightarrow 1 - P\left( X = 0 \right) > 0 . 99\]
\[ \Rightarrow 1 - \frac{1}{4^n} > 0 . 99\]
\[ \Rightarrow \frac{1}{4^n} < 0 . 01\]
\[ \Rightarrow 4^n > \frac{1}{0 . 01}\]
\[ \Rightarrow 4^n > 100\]
\[\text{ The least value of n satisfying this inequality is 4 } . \]
\[\text{ Hence, the shooter must fire at least 4 times . } \]