An archer is shooting at a target with a probability of $0.4$ Find the minimum amount of attempts so the chances to hit the target $80$ times are $0.9$.
I think this can be solved with negative binomal distribution but I'm not sure how. Please help!
What I got so far:
$x:$ the number of experiments $x\: \tilde{}\: NB(80, 0.4)$
$p(x \ge 80) = 0.9$.
It is $P(X\geq 80)=1-P(X\leq 79)$, where $X \sim Bin(n,0.4)$. The equation is
$1-P(X\leq 79)\geq 0.9 \quad |-1$
$-P(X\leq 79)\geq -0.1 \quad |\cdot (-1)$
$P(X\leq 79)\leq 0.1 $
Using De Moivre-Laplace it looks like
$\Phi\left( \frac{79+0.5-n\cdot 0.4}{\sqrt{n\cdot 0.4 \cdot 0.6}} \right)\leq 0.1$
$\Phi(z)$ is the cdf of the standard normal distribution. +0.5 is the continuity correction factor.
$\frac{79+0.5-n\cdot 0.4}{\sqrt{n\cdot 0.4 \cdot 0.6}} \leq \Phi^{-1}\left(0.1\right)$
$\frac{79+0.5-n\cdot 0.4}{\sqrt{n\cdot 0.4 \cdot 0.6}} \leq -1.29$
Now you can solve the (in-)equality for n.
Check the result for n and the check by using the binomial distribution:
Link 1
Link 2