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 X : Number of times he hits the target

Hitting the target is a bernoulli trial.

So, X has binomial distribution

$P(X=x)={n_C}_x{q}^{n-x}{p}^x$

n = number of rounds fired

p = Probability of hitting =$\frac{3}{4}$

q= 1-p=$\frac{1}{4}$

Hence, $P(X=x)={n_C}_x\left(\frac{3}{4}{)}^x\right(\frac{1}{4}{)}^{n-x}$

Given $P(X\ge 1)>99$ %, we need to find n

Now,

$P(X\ge 1)>99$ %

$1-P(X=0)>99$ %

$1-{n_C}_0\left(\frac{3}{4}{)}^0\right(\frac{1}{4}{)}^n>0.99$

$1-(\frac{1}{4}{)}^n>0.99$

$1-0.99=(\frac{1}{4}{)}^n$

$4^n>100$

We know that , $4^4=256$

So, $n\ge 4$

So, the minimum value of n is 4.

So, he must fire at least 4 times.