# Probability question that looks like Bayes Theorm

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

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:

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

• 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 $$\mathsf P(G\mid M)=\dfrac{\mathsf P(M\mid G)~\mathsf P(G)}{\mathsf P(M\mid G)~\mathsf P(G)+\mathsf P(M\mid G^\complement)~\mathsf P(G^\compliment)}$$ – yipz Sep 27 '18 at 12:35
• @yipz : Note that "compliment" and "complement" are two different words. The latter is the one that is appropriate here. – Michael Hardy Sep 27 '18 at 23:11

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 $$\Pr(\text{Mr. X is guilty}\mid \text{match}) \approx \frac 1 3.$$

• 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. – yipz Sep 26 '18 at 23:22
• @Yipz There is no need to further understand the solution: it is wrong. Bayes' Rule is $$\dfrac{\Pr(\text{Guilty and Match})}{\Pr(\text{Guilty and Match})+\Pr(\text{Not Guilty and Match})}$$ – Graham Kemp Sep 27 '18 at 2:19
• @yipz : Probabilities are never independent; rather events or random variables are independent. – Michael Hardy Sep 27 '18 at 2:40
• @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 $\Pr(\text{guilty} \mid \text{match}) = \cdots. \qquad$ – Michael Hardy Sep 27 '18 at 2:50
• @GrahamKemp Or to put it more bluntly: You are wrong. $\qquad$ – Michael Hardy Sep 27 '18 at 2:53