The probability of a shooter hitting a target is $\frac{3}{4}$. How many minimum number of times must he fire so that the probability of hitting the target atleast once is more than 0.99?

4

The correct answer is Option (3) → 4

Given, probability of shooting a target = $p =\frac{3}{4}$,

so, $q = 1-p=1- \frac{3}{4}=\frac{1}{4}$

Let the number of trials be $n$.

Thus, we have a binomial distribution with $p=\frac{3}{4},q=\frac{1}{4}$ and number of trials = $n$.

∴ The probability of hitting the target atleast once = $P(X > 1)$

$=1- P(0)= 1-{^nC}_0 q^n$

$=1-1(\frac{1}{4})^n=1-\frac{1}{4^n}$.

According to given,

$1-\frac{1}{4^n} > 0.99⇒ 1 -\frac{1}{4^n}>\frac{99}{100}$

$⇒ 1-\frac{99}{100}>\frac{1}{4^n}⇒\frac{1}{100}>\frac{1}{4^n}⇒100<4^n$

$⇒4n > 100$, which is satisfied if n is atleast 4.

Hence, the man must shoot the target atleast 4 times.