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?

Answer

N/a

Explanation

Let the shooter fire `n` times, which are the number of trials. Now
`p=3//4` and `q=1-p=1-3/4=1/4`
Now `P(Xge1)gt0.99`
`implies1-P(X=0)gt0.99`
`implies1-.^(n)C_(0).q^(n)gt0.99`
`impliesq^(n)lt0.01`
`implies1/(4^(n))lt0.01`
`implies4^(n)gt1/0.01`
`implies4^(n)gt100`
`implies` Minimum value of `n=4`
Thus the shooter must fire 4 times.