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I am working on a probability question that looks like it has the form of a bayes theorm (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 probabilty MR.X commited the crime

Obviously the probability anyone commited 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

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Let $G$ be the event of Mr X being guilty, and $M$ the event of a DNA match.   You are thereby told the probability that Mr X was the one who committed the crime: $\mathsf P(G)=1/100000$ and the probability for a false positive match: $\mathsf P(M\mid G^\complement)=1/50000$.

You are not told the probability for a true positive match, but assume it is close enough to certainty; that : $\mathsf P(M\mid G)=1$ .

You are asked to find the probability of guilt given a match: $\mathsf P(G\mid M)$. Use Bayes' Rule and the Law of Total Probability:

$$\mathsf P(G\mid M)=\dfrac{\mathsf P(M\mid G)~\mathsf P(G)}{\mathsf P(M\mid G)~\mathsf P(G)+\mathsf P(M\mid G^\complement)~\mathsf P(G)}$$

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  • $\begingroup$ should $$\mathsf P(G\mid M)=\dfrac{\mathsf P(M\mid G)~\mathsf P(G)}{\mathsf P(M\mid G)~\mathsf P(G)+\mathsf P(M\mid G^\complement)~\mathsf P(G)}$$ not be $$\mathsf P(G\mid M)=\dfrac{\mathsf P(M\mid G)~\mathsf P(G)}{\mathsf P(M\mid G)~\mathsf P(G)+\mathsf P(M\mid G^\complement)~\mathsf P(G^\compliment)}$$ $\endgroup$ – yipz Sep 27 '18 at 12:35
  • $\begingroup$ @yipz : Note that "compliment" and "complement" are two different words. The latter is the one that is appropriate here. $\endgroup$ – Michael Hardy Sep 27 '18 at 23:11
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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 $$ \Pr(\text{Mr. X is guilty}\mid \text{match}) \approx \frac 1 3. $$

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  • $\begingroup$ Just to further understand the solution given. The probabilities $\frac{\Pr(\text{match}\mid \text{Mr. X is guilty})}{\Pr(\text{match}\mid\text{Mr. X is not guilty})}$ are independent correct? So we can write $\frac{49999/50000}{1/50000}$ which is what you have given. $\endgroup$ – yipz Sep 26 '18 at 23:22
  • $\begingroup$ @Yipz There is no need to further understand the solution: it is wrong. Bayes' Rule is $$\dfrac{\Pr(\text{Guilty and Match})}{\Pr(\text{Guilty and Match})+\Pr(\text{Not Guilty and Match})}$$ $\endgroup$ – Graham Kemp Sep 27 '18 at 2:19
  • $\begingroup$ @yipz : Probabilities are never independent; rather events or random variables are independent. $\endgroup$ – Michael Hardy Sep 27 '18 at 2:40
  • $\begingroup$ @GrahamKemp : What do you consider to be wrong here? I applied the identity $$ \frac{\Pr(H\mid D)}{\Pr(\text{not H}\mid D)} = \frac{\Pr(H)}{\Pr(\text{not } H)} \times \frac{\Pr(D\mid H)}{\Pr(D\mid \text{not }H)}. $$ This is equivalent to what you stated, except that you omitted to say $\Pr(\text{guilty} \mid \text{match}) = \cdots. \qquad$ $\endgroup$ – Michael Hardy Sep 27 '18 at 2:50
  • $\begingroup$ @GrahamKemp Or to put it more bluntly: You are wrong. $\qquad$ $\endgroup$ – Michael Hardy Sep 27 '18 at 2:53

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