The probability of a shooter hitting a target is `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?

Explanation

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} and q=frac{1}{4}`
such that
`P(X=r)=^{n} C_{r}(frac{3}{4})^{r}(frac{1}{4})^{n} `

`Rightarrow P(X=r)=^{n} C_{r} frac{3^{r}}{4^{n}}`

It is given that` P(X geq 1)>0.99`

`Rightarrow 1-P(X=0)>0.99 `

`Rightarrow 1-frac{1}{4^{n}}>0.99 `

`Rightarrow frac{1}{4^{n}}<0.01 `

`Rightarrow 4^{n}>frac{1}{0.01} `

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