Probability question that looks like Bayes Theorem I am working on a probability question that looks like it has the form of a Bayes theorem (false/positive) however I cannot seem to wrap my head around it
Police have DNA from a criminal that committed a crime. The crime was committed by 1 resident of a population of $100000$. They check the DNA sample against a small random sample of the population. The test only gives a false positive $1$ in $50000$ times. The DNA matches MR.X from the database.
What is the probability MR.X committed the crime?
Obviously the probability anyone committed the crime in the area is 
$$\frac{1}{100000}$$
Is the latter information in the question irrelevant here? Or is there something I am missing on how to use it?

Intuitively, one can say there should be about $2$ false positives among $100\,000$ people, plus one guilty person, so three positives, and one of the three is the culprit.
But more precisely, the expected number of false positives should be $\dfrac 1 {50\,000} \times 99\,999$ rather than $\dfrac 1 {50\,000} \times 100\,000.$ But the difference is of no practical importance.
More precisely:
\begin{align}
& \frac{\Pr(\text{Mr. X is guilty}\mid \text{match})}{\Pr(\text{Mr. X is not guilty}\mid\text{match})} \\[13pt]
= {} & \frac{\Pr(\text{Mr. X is guilty})}{\Pr(\text{Mr. X is not guilty})} \times \frac{\Pr(\text{match}\mid \text{Mr. X is guilty})}{\Pr(\text{match}\mid\text{Mr. X is not guilty})} \\[12pt]
= {} & \frac{1/100\,000}{99\,999/100\,000} \times \frac 1 {1/50\,000} = \frac{50\,000}{99\,999} \approx \frac{50\,000}{100\,000} = \frac 1 2.
\end{align}
Then, since
\begin{align}
& \frac{\Pr(\text{Mr. X is guilty}\mid \text{match})}{\Pr(\text{Mr. X is not guilty}\mid\text{match})} \\[12pt]
= {} & \frac{\Pr(\text{Mr. X is guilty}\mid \text{match})}{1 - \Pr(\text{Mr. X is guilty}\mid\text{match})} \approx \frac 1 2,
\end{align}
we conclude that
$$
\Pr(\text{Mr. X is guilty}\mid \text{match}) \approx \frac 1 3.
$$
