Where is the flaw in my logic in this normal distribution question?

Question

A company makes a large number of steel links for chains. They know that the force required to break any individual link is modelled by a normal distribution with mean 20kN. The company tests chains consisting of four links. If any link breaks, the chain will break. A force of 18kN is applied to all of the chains and 30% break.

Estimate the probability of a single link breaking.

My working:

Let $X$ be the amount of force needed to break a link.

$X - N\left(20\:,\:\sigma ^2\right)$

As we know that $18kN$ of force breaks $30\%$ of chains, then this means that the chance of a link breaking under $18kN$ is $30\%$ (I actually highly suspect that this is wrong but I can't come up with an alternative method).

Using the inverse standard normal distribution, I got the standard deviation to be $3.81$. I was then planning to use this value in a normal distribution calculation to get the probability that only one link breaks, but it's all just wrong wrong wrong wrong.

By the way: I know the actual worked solution to the answer, I just want to know there the problem is.

Answer 1

Since all four links survive with probability $0.7$, each link's individual survival probability is $p:=0.7^{1/4}$, while its breakage probability is $1-p$.

We don't actually need any details of the Normal distribution for the stated problem. Supposing you'd like to think about that, however: working in kN, $\Phi^{-1}(1-p):=\frac{-2}{\sigma}$.

Answer 2

Let $p$ be the probability that a single link breaks. Then, the probability of a chain not breaking has to be \begin{equation} \Pr(\text{Chain doesn't break}) = (1-p)^4 \end{equation} So \begin{equation} \Pr(\text{Chain breaks}) =1-\Pr(\text{Chain doesn't break})= 1- (1-p)^4 \end{equation} You've got $q = 0.3$, then $p=0.085$