Question

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 1

0.0004498

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

`\large P(A^c|B^c)=0.8`

where

`\large  A^c,\;B^c` 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

`\large (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`

Now, we have

`\large P(B|A)=(P(A\cap B))/(P(A))=(0.00009)/(P(A))`

So if we can determine P(A), the result will follow.

By De Morgan's Law

`\large A^c\cap B^c=(A\cup B)^c`

so

`\large 0.8=P(A^c|B^c)= (P(A^c\cap B^c))/(P(B^c))=(P((A\cup B)^c))/(P(B^c))=(1-P(A\cup B))/(1-P(B))\Rightarrow\\\\(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`

Using the formula

`\large P(A\cup B)=P(A)+P(B)-P(A\cap  B)`

and replacing the values we have found

`\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007`

and finally, the desired result is

`\large P(B|A)=(0.00009)/(P(A))=(0.00009)/(0.20007)\Rightarrow\\\\\boxedA)=0.0004498`