9. A judge hears the following arguments in a murder hearing. The DNA test that places the accused at the murder scene has a true positive rate of 90% (i.e. the probability that the test returning positive given that the accused was actually present at the scene is 0.9). Similarly, the DNA test has a false negative rate of 80% (i.e. the probability that the test returns negative given that the accused was not present at the scene is 0.8). Everyone in the town has a equal probability of being at the murder scene, and the town has a population of 10,000. Given the fact that the DNA test returned a positive result for the accused, what is the probability that the accused was at the murder scene?

Answer:0.0004498Step-by-step explanation:Let us define the events: A = The test returns positive. B = The accused was present. Since everyone in the town has an equal probability of being at the murder scene, and the town has a population of 10,000 P(B) = 1/10000 = 0.0001 We have that the probability the test returning positive given that the accused was actually present at the scene is 0.9 P(A | B) = 0.9 and the probability that the test returns negative given that the accused was not present at the scene is 0.8 [tex]\large P(A^c|B^c)=0.8[/tex] where [tex]\large  A^c,\;B^c[/tex] are the complements of A and B respectively. We want to determine the probability that the DNA test returned a positive result given that the accused was at the murder scene, that is, P(B | A). We know that P(A | B) = 0.9, so [tex]\large \frac{P(A\cap B)}{P(B)}=0.9\Rightarrow P(A\cap B)=0.9P(B)=0.9*0.0001\Rightarrow\\\\P(A\cap B)=0.00009[/tex] Now, we have [tex]\large P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.00009}{P(A)}[/tex] So if we can determine P(A), the result will follow. By De Morgan's Law [tex]\large A^c\cap B^c=(A\cup B)^c[/tex] so [tex]\large 0.8=P(A^c|B^c)= \frac{P(A^c\cap B^c)}{P(B^c)}=\frac{P((A\cup B)^c)}{P(B^c)}=\frac{1-P(A\cup B)}{1-P(B)}\Rightarrow\\\\\frac{1-P(A\cup B)}{1-0.0001}=0.8\Rightarrow P(A\cup B)=1-0.8(1-0.0001)\Rightarrow\\\\P(A\cup B)=0.20008[/tex] Using the formula [tex]\large P(A\cup B)=P(A)+P(B)-P(A\cap  B)[/tex] and replacing the values we have found [tex]\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007[/tex] and finally, the desired result is [tex]\large P(B|A)=\frac{0.00009}{P(A)}=\frac{0.00009}{0.20007}\Rightarrow\\\\\boxed{P(B|A)=0.0004498}[/tex]