Binomial Probability Problem Involving Processors Breaking

An electronics company distributes a particular type of processor. When someone orders this particular processor, they are sent a box containing $20$. The company's newest employee, Peter, is tasked with testing the quality of the order before it is dispatched. The company tells Peter there is $1\%$ chanced that a new processor is faulty, therefore, they always keep a box of $5$ spares next to his workstation.

However, someone has accidentally added the wrong processor to the 'replacements box'. We will call this processor, 'processor $A$'.

What is the probability that processor $A$ is added to the box that needs to be shipped? You may assume that each processor is independent of each other processor and that the shipment goes ahead even if there's insufficient processors.

My solution,

Let $X$ be the number of faulty processors out of $20$.

\begin{align} Pr(X \geq 1) & = 1 - Pr(X = 0)\\&=1 - {20\choose 0}(0.01)^0(0.99)^{20}\\ &= 0.82 \end{align}

Let $A$ be processor $A$ is drawn from the spares box,

$Pr(A) = \dfrac{1}{5}$

Now, for the probability that processor is used,

\begin{align} Pr(X \geq 1)\times Pr(A) & = 0.82 \times \dfrac{1}{5} \\ &= 0.16 \end{align}

Therefore, Peter has a $16\%$ chance of adding the wrong processor.

Is this correct? I feel like I can't multiply the probabilities like I have above.

The probabililty that processor $A$ is drawn from the spares box equals to $1/5$ only for the case when there is exactly one wrong processor found. If Peter will find $2$ wrong processors, the probability that $A$ will be drawn from the replacement box along with some other processor is $2/5$. For three wrong processors this probability is $3/5$ and so on. For the case when Peter find $5$ or more wrong processors, $A$ will appear in the box certainly.
So, denote by $A$ the event that processor $A$ is drawn from the spares box. \begin{align} & Pr(A\mid X=1)=\frac15,\\ & Pr(A\mid X=2)=\frac25, \\ & Pr(A\mid X=3)=\frac35,\\ & Pr(A\mid X=4)=\frac45,\\ & Pr(A\mid X\geq 5)=1. \end{align} And $Pr(A\mid X=0)=0$. By Law of Total Probability, $$P(A)=P(X=1)Pr(A\mid X=1)+\ldots+P(X=4)Pr(A\mid X=4)+P(X\geq 5)Pr(A\mid X\geq 5).$$ So you should calculate all the probabilities about $X$ separately and use it in the above formula.