Probability that a patient has Disease $X$ 
Disease $X$ is only present in $0.1$% of patients who are tested. The test is positive $99$% of the time when the patient has Disease $X$. If you are tested for the disease and test positive, then the probability that you have Disease $X$ is $10$%. What is the probability that a person tests positive when they do not have Disease $X$?

What I have tried:
Let $A$ be the probability that the patient has Disease $X$, and $B$ be the probability that they test positive.
Then $P(A)=0.001$, which implies $P(\bar{A})=0.099$ and $\displaystyle P(B/A)=0.99$. Now we have to find $\displaystyle P(B/\bar{A})$.
We also have here: $$P(B)=P(A)P(B/A)+P(\bar{A})P(B/\bar{A}).$$
It seems that we can apply Bayes' theorem. But I do not understand how to apply the formula here.

Using Baye's Theorem, the probability of testing positive is:
\begin{align*}
P(\text{disease} | \text{+test}) =&\ \frac{P(\text{+test} | \text{disease})P(\text{disease})}{P(\text{+test})}\\
P(\text{+test}) =&\ P(\text{+test}|\text{disease})P(\text{disease}) + P(\text{+test} | \text{$\neg$disease})P(\text{$\neg$disease})\\
=&\ .99*0.001 + 0.999x
\end{align*}
We can find $x = P(\text{+test}|\text{$\neg$disease})$ by solving the following equation (I'm mixing percentages with decimals):
\begin{align*}
0.1 = \frac{.99 * 0.1\%}{.99*0.1\% + 99.9\%x}\\
.0099 + 9.99x = .099 \\
x = \frac{0.0891}{9.99} \approx 0.00891891892
\end{align*}
Meaning the probability of a positive test given they do not have the disease is approximately $0.89\%$.
