Conditional probability and criminal DNA analysis

[Moara]

Homework Statement

Two types of DNA test were developed to find the guilty of a crime. For test A, the probability to correctly identify the criminous is $99.5\%$. In $1.5\%$ of cases, test A results in fake-positive (the person is considered guilty, but in fact he is not). For test B, such probabilities are $99.7\%$ and $2\%$, respectively.

The police found a suspect and are $95\%$ sure that he is guilty.

a) Knowing that the test A gave negative, what's the probability that the suspect is guilty?

b) Knowing that both tests gave positive, what's the probability that the suspect is guilty?

c) If the suspect is truly guilty, what's the probability that one test is positive and the other one is negative?

d) Consider that $$10## suspects were caught, and one of them is guilty. What is the probability that test A gives positve only for the guilty suspect?

Relevant Equations

$$P(A|B) = \frac{P(A \ and \ B)}{P(B)}$$

We know that $P(A-) = (95\% \cdot 0.5\% + 5\% \cdot 98.5\% )$ and $P(guilty \ and \ A-) = (95\% \cdot 0.5\%)$, so letter a) is just $P(guilty \ and \ A-)/P(A-)$.

What I tried to do in letter b) was again using the conditional probability theorem. First calculating the probability that both tests give positive

$$P_1=(0.95\cdot 0.995+0.05\cdot 0.015)\cdot (0.95\cdot 0.997+0.05\cdot 0.02)$$

now, intersecting with the event of the suspect being guilty,
$$P_2=0.95\cdot 0.995\cdot 0.95\cdot 0.997$$
$\frac{P_2}{P_1}$ should give the desired result, but it appears that this is not correct, why?

 

[andrewkirk]

Your formula for $P_1$ is wrong. You have multiplied the probabilities of $A^+$ and $B^+$. I think you are assuming that gives the probability of observing both A and B positive. But that's only true if $A^+$ and $B^+$ are independent events. They are not independent, because they both depend on whether the suspect is guilty.

Instead, write probabilities of the mutually exclusive events:
$$
abg=G\wedge A^+\wedge B^+,\ \ \ \ \ \ \ \
abn=(\neg G)\wedge A^+\wedge B^+$$
where $G$ is the event of the suspect being guilty.

Then the conditional probability you seek will be $\frac{abg}{abg+abn}$

Also, what rationale did you have for your formula for $P_2$? Why would you multiply by the probability of being guilty (95%) twice?

Also, note the ambiguity of the question where it says "In 1.5% of cases, test A results in fake-positive". This could mean (a) 1.5% of ALL 'A' tests are false positive, or (b) 1.5% of all 'A' tests of INNOCENT suspects give a positive, or that (c) 1.5% of all positive 'A' results are false-positives. These all give different results.

You have assumed they mean (b) and that seems most likely because that's the approach they used to define the equivalent measure for false negatives.
But it's pretty poor form that they stated the problem in such an ambiguous way.

Finally, why do they assume the suspect is male?