Finding conditional probability and minimum trials in binomial distribution

Question

I have a question like this:

A missile protection system is set up in a particular zone. The system consists of $n$ radar sets operating independently. Each set has a probability of $0.95$ of detecting a missile which enters the zone.

Question

1. Suppose that there are 6 radar sets operating in a particular day (i.e. $n = 6$), Given that a missile is detected by at least one set, what is the conditional probability that it is only detected by exactly one set?

We have: $X \sim B(6, 0.95)$. Therefore: $$P(X = 1 | X \geq 1) = \frac{P(X = 1)}{P(X \geq 1)} = \frac{2}{1122807} \approx 0$$

2. If the probability of detecting a missile in the zone is required to be at least $0.9999$, what is the smallest $n$ can be?

Assuming: $X \sim B(n, 0.95)$. We need: $P(X \geq 1) \geq 0.9999$ or equivalently: $$P(X = 0) \leq 0.0001 = \binom{n}{0}*0.95^{0}*0.05^{n} = 0.05^{n}$$ Therefore, $n \geq 3.0744 $. Minimum $n$ is $4$.

Is my solution right?

Answer 1

Yes, correct. For confirmation, here are computations in R statistical software, where pbinom is a binomial CDF and dbinom is a binomial PDF:

 dbinom(1, 6, .95)/(1 - pbinom(1,6,.95))
 ## 1.781253e-06  #  Very nearly 0

 n = 1:5;  pr = dbinom(0, n, .95)
 cbind(n, pr)
      n        pr
 ##   1 5.000e-02
 ##   2 2.500e-03
 ##   3 1.250e-04
 ##   4 6.250e-06  # <-----
 ##   5 3.125e-07