If hitting a target has $P = 0,3$, how many shoots to get at least one hit with a probability of $0.9$? Cheers, I know that hitting a target has a probability of $0,3$, and I am asked to find the number $n$ of times that I have to shoot at the target to get at least one hit with a probability bigger that $0.9$. I got kinda confused as to which distribution I should be using. because both binomial and geometric seem correct to me.
For example, the geometric will give me the probability of shootings, until I hit the target. But because it talks about at least one hit, I could have more, so I think the binomial is more suitable.
So my try is: let $X$ follow the binomial distribution, out of $n$ trials. I am basically asked to compute $P(x \geq 1) \geq 0.9 \Rightarrow 1 - P(x = 0) \geq 0.9 \Rightarrow P(x = 0) \leq 0.1 \Rightarrow \begin{pmatrix}n \\ 0\end{pmatrix} 0.3^0(0.7)^n \leq 0.1 \Rightarrow n \geq 6.4557$
so I need approximately 6-7 shootings to make sure that I hit at least once with probability of $0.9$ or higher. Would that be correct? Thanks a lot for any help

Assuming independence, the probability that you miss one shot is $0.7$. The probability that you miss both shots out of two is $0.7^2$. The probability that you hit at least one shot out of two shots is $1-0.7^2$ which is the same as $0.3\cdot 0.7\cdot2+0.3^2$ (i.e. hitting exactly one shot or both shots).
The probability of hitting at least one shot out of $n$ shots is therefore $1-0.7^n$. You want this to be higher than $0.9$, so we have $1-0.7^n>0.9$. All that is left is solving this for $n$, which I leave to you.
