A defence system is $99.5\%$ efficient in intercepting ballistic missiles. (a) How many missiles must an aggressor launch to have a better than evens chance of one or more penetrating the defenses?
May you please show some steps to help me solve this? Thank you
If we assume the missiles to represent (independent) Bernoulli trials, the binomial distribution is applicable (since the failure rate $0.005$ is so low, we could approximate this with the Poisson distribution for large $n$).
The probability of $k$ missiles out of $n$ penetrating the defence system is:
$$p(k,n)=\binom{n}{k}(0.005)^k(0.995)^{n-k}$$
For $k=0$ this reduces to
$$p(0,n)=0.995^{n}$$
Hence we are looking for the smallest $n$ for which $0.995^n<0.5$. Then
$$0.995^n=0.5 \implies n\ln(0.995)=\ln(0.5) \implies n=\frac{\ln(0.5)}{\ln(0.995)}\approx138.3$$
So $n=139$ guarantees a better than even chance that at least one missile will get through.