binomial distribution probability - probability for 1 component success out of `n` components

Question

Give the following question:

A missile component have a 5% probability to fail.

In order to enlarge the probability of success of the missile, we install n identical components in that missile, so if one of these n components will success, the missile will success too.

that should n be for a missile success probability greater than 99%?

I just used the [Binomial distribution formula][1]:

$Pr(X = 1) = \binom{n}{1} 0.95^1 0.05^{n-1} = n 0.95^1 0.05^{n-1}$

But as long as n goes higher , $pr(1)$ goes lower.

Where is my problem and how should this question be solved?

Answer 1

$$P(X\geq 1) = \sum_{x=1}^{n} \left(\matrix{n\\x}\right)(1-p)^{x}(p)^{n-x} = 1 - P(X=0)$$ where $X$ is the number of succeses, and $p$ is the probability of failure.

$$ P(X=0) = \left(\matrix{n\\0}\right)(1-p)^{0}p^{n} = p^n $$

Answer 2

The missile succeeds if one or more components succeeds. Let N denote the number of successful components. Then calculate Pr(N >= 1).

Answer 3

We want $P(X\geqslant1)>0.99$. We have that $$P(X\geqslant1)=1-P(X=0)=1-{n \choose 0}(0.95)^0(0.05)^n>0.99 $$This leads to $$1-(0.05)^n>0.99\Rightarrow(0.05)^n<0.01$$This can easily be solved for $n$ using logarithms. Hence we get the result$$n>\frac{log(100)}{log(20)}\simeq1.54$$So the missile will succed for all such values of $n$, and since you can't have $1.54$ of a missile component it's probably best to leave the answer as $n\geqslant2$.